GSEB Class 9 Maths Solutions Chapter 9 Areas of Parallelograms and Triangles Exercise 9.3

Question 1. In the figure, E is any point on the median AD of a ∆ABC. Show that ar(ΔΑΒΕ) = ar(∆ΑΣΕ).
Answer:
Given that E is any point on median AD of a \( \Delta ABC \). In \( \Delta ABC \), AD acts as a median. This means that ar\( (\Delta ABD) = ar(\Delta ACD) \). This is because a median divides a triangle into two triangles that have equal areas.
Similarly, in \( \Delta EBC \), ED serves as a median. Consequently, ar\( (\Delta EBD) = ar(\Delta ECD) \). This follows the same property of a median dividing a triangle into two parts with equal areas.
If we subtract the area of \( \Delta EBD \) from ar\( (\Delta ABD) \), we get ar\( (\Delta ABE) \). Also, if we subtract the area of \( \Delta ECD \) from ar\( (\Delta ACD) \), we get ar\( (\Delta ACE) \). Since ar\( (\Delta ABD) = ar(\Delta ACD) \) and ar\( (\Delta EBD) = ar(\Delta ECD) \), it implies that ar\( (\Delta ABE) = ar(\Delta ACE) \).

The steps are as follows:
Given: E is any point on median AD of a \( \Delta ABC \).
In \( \Delta ABC \), AD is a median.
\( \implies ar(\Delta ABD) = ar(\Delta ACD) \) ..........(1)
(A median of a triangle divides it into two triangles of equal areas)
In \( \Delta EBC \), ED is a median.
\( \implies ar(\Delta EBD) = ar(\Delta ECD) \) ..........(2)
Subtracting (2) from (1), we get
\( ar(\Delta ABD) - ar(\Delta EBD) = ar(\Delta ACD) - ar(\Delta ECD) \)
\( \implies ar(\Delta ABE) = ar(\Delta ACE) \)
In simple words: When a line splits a triangle into two smaller triangles, and that line is a median, then the two smaller triangles will always have the exact same size (area). This rule applies twice here, once for the big triangle and once for a smaller one, which helps us prove that two specific parts of the big triangle have equal areas.

Exam Tip: Remember the key property that a median divides a triangle into two triangles of equal area, and apply it carefully to the correct triangles in the problem.

Question 2. In a triangle ABC, E is the mid-point of median AD. Show that \( ar(\Delta BED) = \frac {1}{4} ar(\Delta ABC) \)
Answer:
In \( \Delta ABC \), AD is a median. This means that it divides the triangle into two parts of equal area, so ar\( (\Delta ABD) = ar(\Delta ACD) \). Each of these areas will be half of the total area of \( \Delta ABC \). So, \( ar(\Delta ABD) = \frac {1}{2} ar(\Delta ABC) \).
Now consider \( \Delta ABD \). E is the mid-point of AD. BE is a median in \( \Delta ABD \). This also means that BE divides \( \Delta ABD \) into two triangles of equal area: ar\( (\Delta ABE) \) and ar\( (\Delta BED) \). Therefore, ar\( (\Delta BED) = \frac {1}{2} ar(\Delta ABD) \).
By substituting the value of ar\( (\Delta ABD) \) from the first step into the second step, we get: \( ar(\Delta BED) = \frac {1}{2} \times \frac {1}{2} ar(\Delta ABC) \).
This calculation simplifies to \( ar(\Delta BED) = \frac {1}{4} ar(\Delta ABC) \).

The detailed steps are:
In \( \Delta ABC \), AD is a median.
\( \implies ar(\Delta ABD) = ar(\Delta ACD) = \frac {1}{2} ar(\Delta ABC) \) ........(1)
(A median of a triangle divides it into two triangles of equal areas)
In \( \Delta ABD \), E is the mid-point of AD.
\( \implies \) BE is a median in \( \Delta ABD \).
So, \( ar(\Delta BED) = \frac {1}{2} ar(\Delta ABD) \)
From (1), substitute \( ar(\Delta ABD) \):
\( ar(\Delta BED) = \frac {1}{2} \times (\frac {1}{2} ar(\Delta ABC)) \)
\( \implies ar(\Delta BED) = \frac {1}{4} ar(\Delta ABC) \)
In simple words: When you have a triangle and a median, the median cuts the triangle exactly in half by area. If you then cut one of those half-triangles in half again with another median, you end up with an area that is a quarter of the original triangle's area.

Exam Tip: Clearly identify the medians and the triangles they divide at each step. This helps in correctly applying the area property of medians sequentially.

Question 3. Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Answer:
Given: ABCD is a parallelogram whose diagonals AC and BD intersect at O. These diagonals divide the parallelogram into four triangles: \( \Delta OAB, \Delta OBC, \Delta OCD \), and \( \Delta ODA \). We need to show that the areas of these four triangles are equal.

In a parallelogram, the diagonals bisect each other. This means that O is the midpoint of both AC and BD. So, \( AO = OC \) and \( BO = OD \).
Consider \( \Delta ABD \). Since O is the midpoint of BD, AO is a median of \( \Delta ABD \). Therefore, ar\( (\Delta AOD) = ar(\Delta AOB) \). (Median property).
Consider \( \Delta ABC \). Since O is the midpoint of AC, BO is a median of \( \Delta ABC \). Therefore, ar\( (\Delta AOB) = ar(\Delta BOC) \).
Consider \( \Delta BCD \). Since O is the midpoint of BD, CO is a median of \( \Delta BCD \). Therefore, ar\( (\Delta BOC) = ar(\Delta COD) \).
Combining these results, we get \( ar(\Delta AOD) = ar(\Delta AOB) = ar(\Delta BOC) = ar(\Delta COD) \). Thus, the diagonals divide the parallelogram into four triangles of equal area.

The detailed proof is:
Given: ABCD is a parallelogram whose diagonals AC and BD intersect at O.
To Prove: \( ar(\Delta OAB) = ar(\Delta OBC) = ar(\Delta OCD) = ar(\Delta ODA) \)
Proof: In parallelogram ABCD, the diagonals bisect each other. Thus, O is the midpoint of AC and also the midpoint of BD.
Consider \( \Delta ABD \). AO is a median of \( \Delta ABD \) (since O is the midpoint of BD).
\( \implies ar(\Delta AOD) = ar(\Delta AOB) \) ..........(1)
Consider \( \Delta ABC \). BO is a median of \( \Delta ABC \) (since O is the midpoint of AC).
\( \implies ar(\Delta AOB) = ar(\Delta BOC) \) ..........(2)
Consider \( \Delta BCD \). CO is a median of \( \Delta BCD \) (since O is the midpoint of BD).
\( \implies ar(\Delta BOC) = ar(\Delta COD) \) ..........(3)
From (1), (2) and (3), we get:
\( ar(\Delta OAB) = ar(\Delta OBC) = ar(\Delta OCD) = ar(\Delta ODA) \).
Alternatively, we can draw a perpendicular from B to AC, say BE. Then:
\( ar(\Delta OAB) = \frac {1}{2} \times OA \times BE \)
\( ar(\Delta OBC) = \frac {1}{2} \times OC \times BE \)
Since \( OA = OC \) (diagonals of a parallelogram bisect each other), then \( ar(\Delta OAB) = ar(\Delta OBC) \) ..........(1)
Similarly, draw a perpendicular from D to AC, say DF. Then \( ar(\Delta ODA) = \frac {1}{2} \times OA \times DF \) and \( ar(\Delta OCD) = \frac {1}{2} \times OC \times DF \). Since \( OA = OC \), then \( ar(\Delta ODA) = ar(\Delta OCD) \) ..........(2)
In \( \Delta AOB \) and \( \Delta COD \):
\( AO = OC \)
\( BO = OD \)
\( \angle AOB = \angle COD \) (vertically opposite angles)
So \( \Delta AOB \cong \Delta COD \) (by SAS congruence criterion).
\( \implies ar(\Delta AOB) = ar(\Delta COD) \) ..........(3)
From (1), (2), and (3), it is clear that all four triangles have equal areas.
In simple words: A parallelogram's diagonals cut each other exactly in the middle. This means each diagonal acts like a median for the two triangles it forms. Because a median divides a triangle into two equal parts area-wise, all four small triangles created by the crossing diagonals end up having the same area.

Exam Tip: Remember that diagonals of a parallelogram bisect each other. This fact is crucial for identifying the medians of the component triangles, which in turn allows you to use the median-area property.

Question 4. In the figure, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O. Show that ar(∆ABC) = ar(ΔΑΒD).
Answer:
Given that triangles ABC and ABD share the same base AB. We also know that line segment CD is bisected by AB at point O. This means that O is the midpoint of CD, so \( OC = OD \).
Consider \( \Delta BCD \). Since O is the midpoint of CD, BO is a median for \( \Delta BCD \). According to the median property, a median divides a triangle into two triangles of equal area. Therefore, \( ar(\Delta OBC) = ar(\Delta OBD) \).
Similarly, consider \( \Delta ACD \). Since O is the midpoint of CD, AO is a median for \( \Delta ACD \). Thus, \( ar(\Delta OAC) = ar(\Delta OAD) \).
Now, we want to prove that \( ar(\Delta ABC) = ar(\Delta ABD) \).
\( ar(\Delta ABC) \) can be written as \( ar(\Delta OAC) + ar(\Delta OBC) \).
\( ar(\Delta ABD) \) can be written as \( ar(\Delta OAD) + ar(\Delta OBD) \).
Since we established that \( ar(\Delta OAC) = ar(\Delta OAD) \) and \( ar(\Delta OBC) = ar(\Delta OBD) \), if we add these two equalities, we get \( ar(\Delta OAC) + ar(\Delta OBC) = ar(\Delta OAD) + ar(\Delta OBD) \).
This directly leads to \( ar(\Delta ABC) = ar(\Delta ABD) \).

The steps are:
Given: Line segment CD is bisected by AB at O.
\( \implies OC = OD \)
In \( \Delta BCD \), BO is a median (since O is the midpoint of CD).
\( \implies ar(\Delta OBC) = ar(\Delta OBD) \) ..........(1)
(A median of a triangle divides it into two triangles of equal areas).
Similarly, in \( \Delta ACD \), AO is a median (since O is the midpoint of CD).
\( \implies ar(\Delta OAC) = ar(\Delta OAD) \) ..........(2)
Adding (1) and (2), we get:
\( ar(\Delta OBC) + ar(\Delta OAC) = ar(\Delta OBD) + ar(\Delta OAD) \)
\( \implies ar(\Delta ABC) = ar(\Delta ABD) \)
In simple words: If a line segment connecting two triangle vertices is cut in half by the base, it means the parts on either side of the base have equal area. By using this idea for both parts of the base, we find that the two triangles that share the base must have the same total area.

Exam Tip: Always look for medians when given information about midpoints or bisection. The property of a median dividing a triangle into equal areas is fundamental here.

Question 5. D, E and F are respectively the mid-points of the sides BC, CA and AB of a ∆ABC. Show that
(i) BDEF is a parallelogram
(ii) \( ar(ADEF) = \frac {1}{4} ar(\Delta ABC) \)
(iii) \( ar(BDEF) = \frac {1}{2} ar(\Delta ABC) \)
Answer:
(i) **To show BDEF is a parallelogram:**
In \( \Delta ABC \), F is the mid-point of side AB and E is the mid-point of side AC. By the Mid-point Theorem, the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is half of its length. Therefore, EF is parallel to BC. Since D is the mid-point of BC, EF is parallel to BD.
Similarly, consider sides AB and BC. F is the mid-point of AB and D is the mid-point of BC. By the Mid-point Theorem, FD is parallel to AC. Since E is the mid-point of AC, FD is parallel to BE.
Since EF \( \parallel \) BD and FD \( \parallel \) BE, the quadrilateral BDEF has both pairs of opposite sides parallel. This proves that BDEF is a parallelogram.

(ii) **To show \( ar(ADEF) = \frac {1}{4} ar(\Delta ABC) \):**
From part (i), we know that BDEF is a parallelogram. Similarly, we can show that AFDE is a parallelogram and CDFE is a parallelogram.
The diagonal of a parallelogram divides it into two triangles of equal area. For parallelogram BDEF, FD is a diagonal, so \( ar(\Delta BFD) = ar(\Delta DEF) \).
For parallelogram AFDE, AD is a diagonal, so \( ar(\Delta AFD) = ar(\Delta AEF) \).
For parallelogram CDFE, CE is a diagonal, so \( ar(\Delta CDE) = ar(\Delta CFE) \).
Also, since F, D, E are mid-points, we can say:
\( ar(\Delta AFE) = ar(\Delta FBD) \) (by congruent triangles, as \( \Delta AFE \cong \Delta BDF \))
\( ar(\Delta AFE) = ar(\Delta EDC) \) (as \( \Delta AFE \cong \Delta CDE \))
\( ar(\Delta FBD) = ar(\Delta EDC) \)
Therefore, all four triangles formed by joining the mid-points ( \( \Delta AFE, \Delta FBD, \Delta EDC, \Delta DEF \) ) have equal areas.
So, \( ar(\Delta AFE) = ar(\Delta FBD) = ar(\Delta EDC) = ar(\Delta DEF) \).
The total area of \( \Delta ABC \) is the sum of these four non-overlapping triangles:
\( ar(\Delta ABC) = ar(\Delta AFE) + ar(\Delta FBD) + ar(\Delta EDC) + ar(\Delta DEF) \)
Since all these areas are equal to \( ar(\Delta DEF) \), we have:
\( ar(\Delta ABC) = 4 \times ar(\Delta DEF) \)
\( \implies ar(\Delta DEF) = \frac {1}{4} ar(\Delta ABC) \). (Note: The question asks for ADEF, which is the same as DEF or FDE, just a different naming convention for the same triangle).

(iii) **To show \( ar(BDEF) = \frac {1}{2} ar(\Delta ABC) \):**
From part (i), BDEF is a parallelogram. Its area is the sum of the areas of two non-overlapping triangles formed by its diagonal FD: \( ar(BDEF) = ar(\Delta BFD) + ar(\Delta DEF) \).
From part (ii), we know that \( ar(\Delta BFD) = ar(\Delta DEF) \).
So, \( ar(BDEF) = ar(\Delta DEF) + ar(\Delta DEF) = 2 \times ar(\Delta DEF) \).
Also from part (ii), we established that \( ar(\Delta DEF) = \frac {1}{4} ar(\Delta ABC) \).
Substituting this into the equation for \( ar(BDEF) \):
\( ar(BDEF) = 2 \times (\frac {1}{4} ar(\Delta ABC)) \)
\( \implies ar(BDEF) = \frac {1}{2} ar(\Delta ABC) \).

Detailed steps:
(i) In \( \Delta ABC \), F is the mid-point of side AB and E is the mid-point of side AC.
By Mid-point theorem, EF \( \parallel \) BC.
Since D is the mid-point of BC, EF \( \parallel \) BD ..........(1)
Similarly, E is the mid-point of AC and D is the mid-point of BC.
By Mid-point theorem, ED \( \parallel \) AB.
Since F is the mid-point of AB, ED \( \parallel \) BF ..........(2)
From (1) and (2), BDEF is a parallelogram.
(A quadrilateral is a parallelogram if its opposite sides are parallel).

(ii) As in (i), we can prove that AFE is a parallelogram. FD is a diagonal of parallelogram BDEF. Similarly, FE is a diagonal of parallelogram CDEF. DE is a diagonal of parallelogram AFED.
The diagonal of a parallelogram divides it into two triangles of equal area.
\( ar(\Delta FBD) = ar(\Delta DEF) \) (from BDEF is a parallelogram)
\( ar(\Delta AFE) = ar(\Delta DEF) \) (from AFED is a parallelogram)
\( ar(\Delta CDE) = ar(\Delta DEF) \) (from CDEF is a parallelogram)
Therefore, \( ar(\Delta AFE) = ar(\Delta FBD) = ar(\Delta EDC) = ar(\Delta DEF) \).
Now, \( \Delta ABC \) is divided into four non-overlapping triangles \( \Delta AFE, \Delta FBD, \Delta EDC \) and \( \Delta DEF \).
\( ar(\Delta ABC) = ar(\Delta AFE) + ar(\Delta FBD) + ar(\Delta EDC) + ar(\Delta DEF) \)
Since all are equal to \( ar(\Delta DEF) \),
\( ar(\Delta ABC) = 4 \times ar(\Delta DEF) \)
\( \implies ar(\Delta DEF) = \frac {1}{4} ar(\Delta ABC) \) ..........(7)

(iii) From (i), BDEF is a parallelogram.
\( ar(BDEF) = ar(\Delta BFD) + ar(\Delta DEF) \)
Since \( ar(\Delta BFD) = ar(\Delta DEF) \),
\( ar(BDEF) = ar(\Delta DEF) + ar(\Delta DEF) = 2 \times ar(\Delta DEF) \)
From (7), \( ar(\Delta DEF) = \frac {1}{4} ar(\Delta ABC) \)
\( ar(BDEF) = 2 \times (\frac {1}{4} ar(\Delta ABC)) \)
\( \implies ar(BDEF) = \frac {1}{2} ar(\Delta ABC) \)
In simple words: First, if you connect the middle points of the sides of a triangle, you create a parallelogram. Second, this parallelogram takes up exactly half the area of the original triangle. Third, the central small triangle (formed by the three midpoints) has an area that is one-fourth of the big triangle's area.

Exam Tip: The Mid-point Theorem is essential here for proving parallelograms. Also, remember that a diagonal divides a parallelogram into two triangles of equal area, and that forming a triangle by joining the midpoints results in four congruent (and thus equal area) triangles.

Question 6. In figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
(i) ar(ADOC) = ar(ΔΑΟΒ)
(ii) ar(ADCB) = ar(ΔΑCB)
(iii) DA || CB or ABCD is a parallelogram
[Hint: From D and B, draw perpendiculars to AC.]
Answer:
Construction: Draw DE \( \perp \) AC and BF \( \perp \) AC.

(i) **To show \( ar(\Delta DOC) = ar(\Delta AOB) \):**
Consider \( \Delta DOE \) and \( \Delta BOF \).
\( \angle DEO = \angle BFO = 90^\circ \) (by construction, DE \( \perp \) AC and BF \( \perp \) AC).
\( \angle DOE = \angle BOF \) (vertically opposite angles).
\( OD = OB \) (given).
Therefore, \( \Delta DOE \cong \Delta BOF \) by the AAS congruence rule.
Since the triangles are congruent, their areas are equal: \( ar(\Delta DOE) = ar(\Delta BOF) \).
Also, corresponding parts of congruent triangles are equal, so \( DE = BF \).

Now consider \( \Delta DEC \) and \( \Delta BFA \).
\( \angle DEC = \angle BFA = 90^\circ \) (by construction).
Hypotenuse \( CD = AB \) (given).
Side \( DE = BF \) (proved above).
Therefore, \( \Delta DEC \cong \Delta BFA \) by the RHS (Right angle-Hypotenuse-Side) congruence rule.
Since the triangles are congruent, their areas are equal: \( ar(\Delta DEC) = ar(\Delta BFA) \).

Add \( ar(\Delta DOE) \) to \( ar(\Delta DEC) \) and \( ar(\Delta BOF) \) to \( ar(\Delta BFA) \).
\( ar(\Delta DOE) + ar(\Delta DEC) = ar(\Delta BOF) + ar(\Delta BFA) \)
\( \implies ar(\Delta DOC) = ar(\Delta AOB) \).

(ii) **To show \( ar(\Delta DCB) = ar(\Delta ACB) \):**
We know from part (i) that \( ar(\Delta DOC) = ar(\Delta AOB) \).
Add \( ar(\Delta OBC) \) to both sides:
\( ar(\Delta DOC) + ar(\Delta OBC) = ar(\Delta AOB) + ar(\Delta OBC) \)
\( \implies ar(\Delta DCB) = ar(\Delta ACB) \).

(iii) **To show DA \( \parallel \) CB or ABCD is a parallelogram:**
From part (ii), \( ar(\Delta DCB) = ar(\Delta ACB) \).
These two triangles, \( \Delta DCB \) and \( \Delta ACB \), have the same base BC.
If two triangles have the same base and equal areas, then they must lie between the same pair of parallel lines. Therefore, DA must be parallel to CB.
A quadrilateral is a parallelogram if both pairs of opposite sides are parallel. We already have AB = CD (given) and DA \( \parallel \) CB (proved above).
Also, from part (i), we proved \( \Delta DOE \cong \Delta BOF \), which implies \( OE = OF \).
From \( \Delta DEC \cong \Delta BFA \), we get \( CE = AF \).
Adding \( OE = OF \) and \( CE = AF \):
\( OE + CE = OF + AF \)
\( \implies OC = OA \).
Since the diagonals AC and BD bisect each other (because \( OA = OC \) and \( OB = OD \)), ABCD is a parallelogram.

Detailed steps:
Construction: Draw DE \( \perp \) AC and BF \( \perp \) AC.
(i) In \( \Delta DOE \) and \( \Delta BOF \):
\( \angle DEO = \angle BFO \) (each \( 90^\circ \))
\( \angle DOE = \angle BOF \) (Vertically opposite angles)
\( OB = OD \) (given)
\( \implies \Delta DOE \cong \Delta BOF \) (By AAS congruency rule)
\( \implies DE = BF \) and \( OE = OF \) ..........(1) (CPCT)
Also, \( ar(\Delta DOE) = ar(\Delta BOF) \) ..........(2)
Now, in right \( \Delta DEC \) and \( \Delta BFA \):
Hypotenuse \( DC = AB \) (Given)
Side \( DE = BF \) (Proved above)
\( \angle DEC = \angle BFA \) (each \( 90^\circ \))
\( \implies \Delta DEC \cong \Delta BFA \) (By RHS rule)
\( \implies ar(\Delta DEC) = ar(\Delta BFA) \) ..........(3)
Adding (2) and (3):
\( ar(\Delta DOE) + ar(\Delta DEC) = ar(\Delta BOF) + ar(\Delta BFA) \)
\( \implies ar(\Delta DOC) = ar(\Delta AOB) \)

(ii) We have \( ar(\Delta DOC) = ar(\Delta AOB) \)
Adding \( ar(\Delta OBC) \) to both sides:
\( ar(\Delta DOC) + ar(\Delta OBC) = ar(\Delta AOB) + ar(\Delta OBC) \)
\( \implies ar(\Delta DCB) = ar(\Delta ACB) \)

(iii) Since \( ar(\Delta DCB) = ar(\Delta ACB) \), and both triangles lie on the same base BC, their corresponding altitudes must be equal. This implies that they lie between the same parallel lines. Thus, DA \( \parallel \) CB.
We are given AB = CD, and we have proven DA \( \parallel \) CB. If one pair of opposite sides is parallel and equal, the quadrilateral is a parallelogram.
Alternatively, we found \( OA = OC \) and \( OB = OD \) (from previous congruent triangles and given). Since the diagonals bisect each other, ABCD is a parallelogram. Therefore, DA \( \parallel \) CB.
In simple words: This problem shows that if two triangles on the same base have equal areas, their vertices must lie on a line parallel to the base. By breaking down the quadrilateral into smaller triangles and using congruence and area properties, we can show that opposite sides are parallel, making the figure a parallelogram.

Exam Tip: This question is a classic proof. Remember to use construction (perpendiculars) to establish congruence between triangles, which helps in proving area equality and ultimately parallelism. The AAS and RHS congruence rules are key here.

Question 7. D and E are points on sides AB and AC respectively of ∆ABC such that ar(ADBC) = ar(ΔΕΒC). Prove that DE || BC.
Answer:
Given that D and E are points on sides AB and AC, respectively, of \( \Delta ABC \). We are provided with the information that the area of \( \Delta DBC \) is equal to the area of \( \Delta EBC \), i.e., \( ar(\Delta DBC) = ar(\Delta EBC) \).
We need to prove that the line segment DE is parallel to BC.
Observe that both \( \Delta DBC \) and \( \Delta EBC \) share a common base, which is BC. When two triangles have the same base and also have equal areas, a fundamental property of geometry states that their vertices not on the common base must lie on a line parallel to the common base. In this case, the vertices not on BC are D and E.
Since \( ar(\Delta DBC) = ar(\Delta EBC) \) and they share the base BC, it logically follows that the line segment DE, connecting the vertices D and E, must be parallel to the common base BC.
Thus, DE \( \parallel \) BC.
In simple words: If two triangles sitting on the same base have the same size (area), then the line that connects their top points must be flat and parallel to their shared base.

Exam Tip: The key theorem to remember is: "Two triangles having the same base (or equal bases) and equal areas lie between the same parallels." This is a direct application of that theorem.

Question 8. XY is a line parallel to BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that ar(ΔΑΒΕ) = ar(ΔΑCF).
Answer:
Given that XY is a line parallel to BC. Also, BE \( \parallel \) AC and CF \( \parallel \) AB. These lines BE and CF meet XY at points E and F, respectively. We need to prove that \( ar(\Delta ABE) = ar(\Delta ACF) \).

Consider the quadrilateral BCFX. We are given that XY \( \parallel \) BC, which means XF \( \parallel \) BC. Also, we are given CF \( \parallel \) AB, which implies CF \( \parallel \) BX. Since both pairs of opposite sides are parallel, BCFX is a parallelogram.
In a parallelogram, opposite sides are equal. So, BC = XF.

Now consider quadrilateral BCYE. We are given that XY \( \parallel \) BC, so EY \( \parallel \) BC. We are also given BE \( \parallel \) AC, which means BE \( \parallel \) CY. Since both pairs of opposite sides are parallel, BCYE is a parallelogram.
In a parallelogram, opposite sides are equal. So, BC = EY.

From BC = XF and BC = EY, we can conclude that XF = EY. Since XF and EY are parts of the same line XY, and they meet at Y, it means YF = XE.

Consider \( \Delta AEX \) and \( \Delta AFY \). They have bases XE and YF, which we just showed are equal ( \( XE = YF \) ). Also, they share a common vertex A, and since XY \( \parallel \) BC, the altitude from A to XY is the same for both triangles. Triangles with equal bases and the same altitude have equal areas. Therefore, \( ar(\Delta AEX) = ar(\Delta AFY) \).

Now consider \( \Delta BEX \) and \( \Delta CFY \). These triangles have bases XE and YF respectively. Since XE = YF, their bases are equal. Also, they lie between the same parallels EF (which is part of XY) and BC (since XY \( \parallel \) BC). Triangles with equal bases and lying between the same parallels have equal areas. So, \( ar(\Delta BEX) = ar(\Delta CFY) \).

Finally, add the corresponding areas:
\( ar(\Delta AEX) + ar(\Delta BEX) = ar(\Delta AFY) + ar(\Delta CFY) \)
\( \implies ar(\Delta ABE) = ar(\Delta ACF) \).

Detailed steps:
Given: XY \( \parallel \) BC. BE \( \parallel \) AC. CF \( \parallel \) AB.
In quadrilateral BCFX:
XF \( \parallel \) BC (since XY \( \parallel \) BC is given)
CF \( \parallel \) BX (since CF \( \parallel \) AB is given)
\( \implies \) BCFX is a parallelogram.
Thus, BC = XF (opposite sides of a parallelogram) ..........(1)
In quadrilateral BCYE:
EY \( \parallel \) BC (since XY \( \parallel \) BC is given)
BE \( \parallel \) CY (since BE \( \parallel \) AC is given)
\( \implies \) BCYE is a parallelogram.
Thus, BC = EY (opposite sides of a parallelogram) ..........(2)
From (1) and (2), we get XF = EY.
Now, \( XY + YF = XY + XE \)
\( \implies YF = XE \) ..........(3)

Consider \( \Delta AEX \) and \( \Delta AFY \).
They have equal bases (XE = YF from (3)) on the same line EF and have a common vertex A.
Therefore, their altitudes from A to EF are also the same.
\( \implies ar(\Delta AEX) = ar(\Delta AFY) \) ..........(4)

Consider \( \Delta BEX \) and \( \Delta CFY \).
They have equal bases (XE = YF from (3)) on the same line EF and are between the same parallels EF and BC (since XY \( \parallel \) BC).
\( \implies ar(\Delta BEX) = ar(\Delta CFY) \) ..........(5)
(Two triangles on the same base (or equal bases) and between the same parallels are equal in area).

Adding (4) and (5), we get:
\( ar(\Delta AEX) + ar(\Delta BEX) = ar(\Delta AFY) + ar(\Delta CFY) \)
\( \implies ar(\Delta ABE) = ar(\Delta ACF) \)
In simple words: This proof uses two main ideas: first, identifying parallelograms formed by the given parallel lines to show that certain line segments are equal. Second, using the property that triangles with the same base (or equal bases) and between the same parallel lines have equal areas. By combining these, we prove the two main triangles have equal areas.

Exam Tip: Break down the problem into smaller quadrilaterals and triangles. Use the properties of parallelograms (opposite sides parallel and equal) and the area theorem for triangles between parallel lines. Carefully identify the bases and parallel lines for each pair of triangles.

Question 9. The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed. Show that ar(|| gm ABCD) = ar(|| gm PBQR)
Answer:
To Prove: \( ar(||gm \ ABCD) = ar(||gm \ PBQR) \)
Construction: Join AC and PQ.

Proof: In parallelogram ABCD, AC is a diagonal. A diagonal divides a parallelogram into two triangles of equal area. Therefore, \( ar(\Delta ABC) = \frac {1}{2} ar(||gm \ ABCD) \). This can be labeled as equation (1).
In parallelogram PBQR, PQ is a diagonal. Similarly, a diagonal divides a parallelogram into two triangles of equal area. Therefore, \( ar(\Delta PBQ) = \frac {1}{2} ar(||gm \ PBQR) \). This can be labeled as equation (2).

Next, consider \( \Delta ACQ \) and \( \Delta APQ \). These two triangles are on the same base AQ. We are given that a line through A is parallel to CP, which means AQ \( \parallel \) CP. Since \( \Delta ACQ \) and \( \Delta APQ \) share the same base AQ and lie between the same parallel lines AQ and CP, their areas must be equal. So, \( ar(\Delta ACQ) = ar(\Delta APQ) \).

Now, subtract the area of \( \Delta ABQ \) from both sides of the equation \( ar(\Delta ACQ) = ar(\Delta APQ) \).
\( ar(\Delta ACQ) - ar(\Delta ABQ) = ar(\Delta APQ) - ar(\Delta ABQ) \)
This simplifies to \( ar(\Delta ABC) = ar(\Delta PBQ) \).

From equation (1), we have \( ar(\Delta ABC) = \frac {1}{2} ar(||gm \ ABCD) \).
From equation (2), we have \( ar(\Delta PBQ) = \frac {1}{2} ar(||gm \ PBQR) \).
Since \( ar(\Delta ABC) = ar(\Delta PBQ) \), we can substitute these values:
\( \frac {1}{2} ar(||gm \ ABCD) = \frac {1}{2} ar(||gm \ PBQR) \)
Multiplying both sides by 2, we get:
\( ar(||gm \ ABCD) = ar(||gm \ PBQR) \).

Detailed steps:
To Prove: \( ar(||gm \ ABCD) = ar(||gm \ PBQR) \)
Construction: Join AC and PQ.
Proof: In parallelogram ABCD, AC is a diagonal.
\( \implies ar(\Delta ABC) = \frac {1}{2} ar(||gm \ ABCD) \) ........(1)
In parallelogram PBQR, PQ is a diagonal.
\( \implies ar(\Delta PBQ) = \frac {1}{2} ar(||gm \ PBQR) \) ........(2)
Now, AQ \( \parallel \) CP (given that a line through A parallel to CP meets CB produced at Q, so AQ is parallel to CP).
Consider \( \Delta ACQ \) and \( \Delta APQ \). They are on the same base AQ and between the same parallels AQ and CP.
\( \implies ar(\Delta ACQ) = ar(\Delta APQ) \)
Subtract \( ar(\Delta ABQ) \) from both sides:
\( ar(\Delta ACQ) - ar(\Delta ABQ) = ar(\Delta APQ) - ar(\Delta ABQ) \)
\( \implies ar(\Delta ABC) = ar(\Delta PBQ) \)
From (1) and (2), we can write:
\( \frac {1}{2} ar(||gm \ ABCD) = \frac {1}{2} ar(||gm \ PBQR) \)
\( \implies ar(||gm \ ABCD) = ar(||gm \ PBQR) \)
In simple words: We want to show two parallelograms have the same area. First, we relate their areas to the areas of triangles inside them. Then, we find two other triangles that share a base and are between parallel lines, proving they have equal areas. By subtracting a common area from these, we show the initial pair of triangles have equal areas, which in turn means the parallelograms have equal areas.

Exam Tip: This proof relies on two main area theorems: 1) A diagonal divides a parallelogram into two triangles of equal area, and 2) Triangles on the same base and between the same parallel lines have equal areas. Pay close attention to identifying the correct bases and parallel lines for each step.

Question 10. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar(ΔΑΟD) = ar(ΔΒΟC)
Answer:
Given that ABCD is a trapezium with AB \( \parallel \) DC. Its diagonals AC and BD intersect at point O. We need to prove that \( ar(\Delta AOD) = ar(\Delta BOC) \).

Consider \( \Delta ABD \) and \( \Delta ABC \). Both of these triangles share the same base AB. Since AB \( \parallel \) DC, it means that AB and DC are parallel lines. Therefore, \( \Delta ABD \) and \( \Delta ABC \) lie between the same parallel lines AB and DC.
A property of triangles states that two triangles on the same base and between the same parallel lines have equal areas. So, \( ar(\Delta ABD) = ar(\Delta ABC) \).

Now, let's look at the areas we want to compare. We can express \( ar(\Delta ABD) \) as the sum of \( ar(\Delta AOB) \) and \( ar(\Delta AOD) \). Similarly, \( ar(\Delta ABC) \) can be expressed as the sum of \( ar(\Delta AOB) \) and \( ar(\Delta BOC) \).
So, we have the equation: \( ar(\Delta AOB) + ar(\Delta AOD) = ar(\Delta AOB) + ar(\Delta BOC) \).
Notice that \( ar(\Delta AOB) \) is common to both sides of the equation. We can subtract \( ar(\Delta AOB) \) from both sides without changing the equality.
Subtracting \( ar(\Delta AOB) \) from both sides, we get:
\( ar(\Delta AOD) = ar(\Delta BOC) \).
This proves the required statement.
In simple words: In a shape with two parallel sides (a trapezium), if two triangles share one of these parallel sides as a base, they will have the same area. When you then subtract the middle overlapping part from both these equal triangles, the remaining non-overlapping parts (on either side of the parallel base) will also have equal areas.

Exam Tip: The core idea is to identify triangles that share a common base and lie between parallel lines. This allows you to state that their areas are equal, and then by subtracting a common overlapping area, you can prove the equality of the desired non-overlapping parts.

Question 11. In figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that:
(i) ar(AACB) = ar(AACF)
(ii) ar(□AEDF) = ar(ABCDE)
Answer:
(i) **To show \( ar(\Delta ACB) = ar(\Delta ACF) \):**
Consider \( \Delta ACB \) and \( \Delta ACF \). Both of these triangles share the same base AC. We are given that a line through B is parallel to AC, and this line meets DC produced at F. This means that BF is parallel to AC.
Since \( \Delta ACB \) and \( \Delta ACF \) are on the same base AC and lie between the same parallel lines AC and BF, their areas must be equal.
Therefore, \( ar(\Delta ACB) = ar(\Delta ACF) \).

(ii) **To show \( ar(Quadrilateral \ AEDF) = ar(Pentagon \ ABCDE) \):**
From part (i), we have established that \( ar(\Delta ACB) = ar(\Delta ACF) \).
Now, let's consider the area of the pentagon ABCDE. It can be seen as the sum of the area of \( \Delta ACB \) and the area of the quadrilateral AEDC. That is, \( ar(ABCDE) = ar(\Delta ACB) + ar(Quadrilateral \ AEDC) \).
Similarly, let's consider the area of the quadrilateral AEDF. It can be seen as the sum of the area of \( \Delta ACF \) and the area of the quadrilateral AEDC. That is, \( ar(AEDF) = ar(\Delta ACF) + ar(Quadrilateral \ AEDC) \).
Since we know that \( ar(\Delta ACB) = ar(\Delta ACF) \), we can substitute \( ar(\Delta ACF) \) for \( ar(\Delta ACB) \) in the pentagon's area equation:
\( ar(ABCDE) = ar(\Delta ACF) + ar(Quadrilateral \ AEDC) \).
Comparing this with the area of quadrilateral AEDF, we find that both expressions are identical.
Therefore, \( ar(Quadrilateral \ AEDF) = ar(Pentagon \ ABCDE) \).

Detailed steps:
(i) In \( \Delta ACB \) and \( \Delta ACF \):
Both triangles are on the same base AC.
It is given that a line through B is parallel to AC (so AC \( \parallel \) BF).
\( \implies \Delta ACB \) and \( \Delta ACF \) are between the same parallels AC and BF.
Therefore, \( ar(\Delta ACB) = ar(\Delta ACF) \).
(Two triangles on the same base (or equal bases) and between the same parallels are equal in area).

(ii) From (i), \( ar(\Delta ACB) = ar(\Delta ACF) \).
Add \( ar(Quadrilateral \ AEDC) \) to both sides:
\( ar(\Delta ACB) + ar(Quadrilateral \ AEDC) = ar(\Delta ACF) + ar(Quadrilateral \ AEDC) \)
The left side represents the area of the pentagon ABCDE.
The right side represents the area of the quadrilateral AEDF.
\( \implies ar(Pentagon \ ABCDE) = ar(Quadrilateral \ AEDF) \).
In simple words: If two triangles share a base and lie between parallel lines, their areas are the same. This lets us swap one triangle's area for another. When we swap an area in a bigger shape (like a pentagon), we can turn it into a different shape (like a quadrilateral) that has the exact same total area.

Exam Tip: This problem demonstrates how to transform the area of a polygon by replacing a triangular part with another triangle of equal area. This is a common technique in area-related proofs, often used to simplify complex shapes into simpler equivalent areas.

Question 12. A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given an equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
Answer:
Let ABCD be the initial quadrilateral plot of land belonging to Itwaari. The Gram Panchayat wants to take over a portion of his land, say \( \Delta ADE \), from one corner (D) to build a Health Centre. Itwaari agrees, provided he gets an equal amount of land (with the same area) adjoining his plot, so his total remaining land becomes a single triangular plot.

Here's how this proposal can be implemented:
1. **Identify the section to be taken:** Let the portion of land to be taken from corner D be represented by \( \Delta ADE \). 2. **Draw a diagonal:** Join the diagonal AC of the quadrilateral ABCD. 3. **Draw a parallel line:** From point D, draw a line parallel to AC. Let this line intersect the extended side BC at point P. 4. **Identify the new land:** The land \( \Delta PCE \) (or \( \Delta EPC \)) is the land that Itwaari will receive. 5. **Form the new plot:** Itwaari's new combined plot of land will be \( \Delta ABP \).

**Proof that the areas are equal:**
Consider \( \Delta DAC \) and \( \Delta PAC \). They are on the same base AC. Since we constructed DP \( \parallel \) AC, these two triangles lie between the same parallel lines AC and DP. Therefore, their areas are equal: \( ar(\Delta DAC) = ar(\Delta PAC) \).
Now, the area that Panchayat takes is \( ar(\Delta ADE) \). The land Itwaari gets in return is \( ar(\Delta EPC) \).
We know \( ar(\Delta DAC) = ar(\Delta PAC) \).
We can write \( ar(\Delta DAC) = ar(\Delta ADE) + ar(\Delta AEC) \).
And \( ar(\Delta PAC) = ar(\Delta EPC) + ar(\Delta AEC) \).
Substituting these into \( ar(\Delta DAC) = ar(\Delta PAC) \):
\( ar(\Delta ADE) + ar(\Delta AEC) = ar(\Delta EPC) + ar(\Delta AEC) \).
Subtracting \( ar(\Delta AEC) \) from both sides, we get:
\( ar(\Delta ADE) = ar(\Delta EPC) \).
This means the area of land taken from Itwaari ( \( \Delta ADE \) ) is equal to the area of land given to him ( \( \Delta EPC \) ).

**Itwaari's new plot:** His original land was quadrilateral ABCD. The Panchayat takes \( \Delta ADE \) (which is part of \( \Delta ADC \)). He is given \( \Delta EPC \).
His total new land becomes \( ar(ABCE) + ar(\Delta EPC) \). Since \( ar(\Delta EPC) = ar(\Delta ADE) \), his new land area is \( ar(ABCE) + ar(\Delta ADE) \), which is precisely \( ar(ABCD) \). Also, this new shape \( ABCE + \Delta EPC \) forms a single triangular plot \( \Delta ABP \).
So, Itwaari gets a new plot \( \Delta ABP \) whose area is equal to the original quadrilateral ABCD. The portion \( \Delta ADE \) is taken, and \( \Delta EPC \) is given. The net effect is a transformation of his plot into a triangular shape without losing any area.

Detailed steps:
Let ABCD be the plot of land of the shape of a quadrilateral. Let the portion \( \Delta ADE \) be taken over by the Gram Panchayat for a Health Centre.
1. Join AC.
2. Draw a line through D parallel to AC to meet BC produced at P.
3. Then Itwaari must be given the land \( \Delta EPC \) adjoining his plot so as to form a triangular plot \( \Delta ABP \).

Proof:
Consider \( \Delta DAC \) and \( \Delta PAC \). They are on the same base AC and between the parallels DP and AC.
\( \implies ar(\Delta DAC) = ar(\Delta PAC) \). (Two triangles on the same base (or equal bases) and between the same parallels are equal in area.)
Subtract \( ar(\Delta AEC) \) from both sides:
\( ar(\Delta DAC) - ar(\Delta AEC) = ar(\Delta PAC) - ar(\Delta AEC) \)
\( \implies ar(\Delta ADE) = ar(\Delta EPC) \)
Thus, the area of land taken from Itwaari ( \( \Delta ADE \) ) is equal to the area of land given to him ( \( \Delta EPC \) ).
Itwaari's new plot area is \( ar(ABCE) + ar(\Delta EPC) \).
Since \( ar(\Delta EPC) = ar(\Delta ADE) \), the new plot area is \( ar(ABCE) + ar(\Delta ADE) \), which is equivalent to the original \( ar(ABCD) \). This new plot forms a single triangle \( \Delta ABP \).
In simple words: To swap a corner of a four-sided plot for an equal amount of land to make a three-sided plot, first draw a diagonal from the corner to be removed. Then, draw a line parallel to this diagonal from the corner point until it meets the extended adjacent side. The new triangle formed by the parallel line and the extended side will have the same area as the original corner piece, and the total land will become a single triangle.

Exam Tip: This is a practical application of the area theorem: "Two triangles on the same base and between the same parallel lines have equal areas." The key steps involve drawing the correct parallel line and recognizing that subtracting a common area from two equal triangles leaves two other equal areas.

Question 13. ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar(AADX) = ar(ΔΑCY).
[Hint: Join CX]
Answer:
Given that ABCD is a trapezium with AB \( \parallel \) DC. A line XY is drawn such that XY \( \parallel \) AC, where X is on AB and Y is on BC.
We need to prove that \( ar(\Delta ADX) = ar(\Delta ACY) \).
Hint: Join CX.

First, consider \( \Delta ADX \) and \( \Delta ACX \). They are on the same base AX. We know that AB \( \parallel \) DC. Since X is on AB and D and C are part of the line DC, it means AX \( \parallel \) DC. Therefore, \( \Delta ADX \) and \( \Delta ACX \) lie between the same parallel lines AX and DC.
So, \( ar(\Delta ADX) = ar(\Delta ACX) \). This can be labeled as equation (1).

Next, consider \( \Delta ACX \) and \( \Delta ACY \). They are on the same base AC. We are given that XY \( \parallel \) AC. Therefore, \( \Delta ACX \) and \( \Delta ACY \) lie between the same parallel lines AC and XY.
So, \( ar(\Delta ACX) = ar(\Delta ACY) \). This can be labeled as equation (2).

From (1) and (2), we have:
\( ar(\Delta ADX) = ar(\Delta ACX) \) and \( ar(\Delta ACX) = ar(\Delta ACY) \).
By transitivity, if the first is equal to the second, and the second is equal to the third, then the first must be equal to the third.
Therefore, \( ar(\Delta ADX) = ar(\Delta ACY) \).
This completes the proof.
In simple words: This problem uses the core idea that triangles sharing a base and positioned between parallel lines have equal areas. We use this rule twice: first, to show that a triangle on one side of the trapezium has the same area as a related triangle, and then again to connect that area to the final target triangle, proving they are all equal in area.

Exam Tip: The key to solving this type of problem is to strategically apply the theorem "Triangles on the same base (or equal bases) and between the same parallels are equal in area" multiple times. Identify common bases and parallel lines clearly.

Question 14. In figure, AP || BQ || CR. Prove that ar(∆AQC) = ar(APBR).
Answer:
Given that lines AP, BQ, and CR are all parallel to each other (AP \( \parallel \) BQ \( \parallel \) CR). We need to prove that the area of \( \Delta AQC \) is equal to the area of the quadrilateral APBR.

Consider \( \Delta BQA \) and \( \Delta BQP \). Both triangles share the same base BQ. Since AP \( \parallel \) BQ (given), these two triangles lie between the same parallel lines BQ and AP. Therefore, their areas are equal: \( ar(\Delta BQA) = ar(\Delta BQP) \). This can be labeled as equation (1).

Now, consider \( \Delta BQC \) and \( \Delta BQR \). Both triangles share the same base BQ. Since BQ \( \parallel \) CR (given), these two triangles lie between the same parallel lines BQ and CR. Therefore, their areas are equal: \( ar(\Delta BQC) = ar(\Delta BQR) \). This can be labeled as equation (2).

We want to find \( ar(\Delta AQC) \). We can write \( ar(\Delta AQC) \) as the sum of \( ar(\Delta BQA) \) and \( ar(\Delta BQC) \).
So, \( ar(\Delta AQC) = ar(\Delta BQA) + ar(\Delta BQC) \).

We also want to find \( ar(APBR) \). We can write \( ar(APBR) \) as the sum of \( ar(\Delta BQP) \) and \( ar(\Delta BQR) \).
So, \( ar(APBR) = ar(\Delta BQP) + ar(\Delta BQR) \).

Now, substitute the equalities from (1) and (2) into the equation for \( ar(APBR) \):
Since \( ar(\Delta BQP) = ar(\Delta BQA) \) and \( ar(\Delta BQR) = ar(\Delta BQC) \),
\( ar(APBR) = ar(\Delta BQA) + ar(\Delta BQC) \).
Comparing this with the expression for \( ar(\Delta AQC) \), we see that they are identical.
Therefore, \( ar(\Delta AQC) = ar(APBR) \).
In simple words: When you have three parallel lines, you can form pairs of triangles that share a base and are between parallel lines, making their areas equal. By carefully adding these equal areas, we can show that the area of a larger triangle is the same as the area of a quadrilateral formed by those lines.

Exam Tip: The key insight is to break down the complex area (quadrilateral APBR and triangle AQC) into sums of simpler triangles. Then, use the fundamental theorem about triangles on the same base and between parallel lines to establish the equality of corresponding parts, allowing you to sum them up for the final proof.

Question 15. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(ΔΑΟD) = ar(ΔΒΟC). Prove that ABCD is a trapezium.
Answer:
Given that ABCD is a quadrilateral whose diagonals AC and BD intersect at O. We are told that \( ar(\Delta AOD) = ar(\Delta BOC) \). We need to prove that ABCD is a trapezium.

Start with the given information: \( ar(\Delta AOD) = ar(\Delta BOC) \).
Add \( ar(\Delta AOB) \) to both sides of this equation:
\( ar(\Delta AOD) + ar(\Delta AOB) = ar(\Delta BOC) + ar(\Delta AOB) \).

The left side of the equation, \( ar(\Delta AOD) + ar(\Delta AOB) \), represents the total area of \( \Delta ABD \).
The right side of the equation, \( ar(\Delta BOC) + ar(\Delta AOB) \), represents the total area of \( \Delta ABC \).
So, we have \( ar(\Delta ABD) = ar(\Delta ABC) \).

Now, consider \( \Delta ABD \) and \( \Delta ABC \). Both of these triangles share a common base, which is AB. If two triangles have the same base and their areas are equal, then their vertices not on the common base must lie on a line parallel to the common base. In this case, the vertices not on AB are D and C.
Therefore, the line segment DC must be parallel to the line segment AB. That is, DC \( \parallel \) AB.
A quadrilateral is defined as a trapezium if exactly one pair of its opposite sides is parallel.
Since we have proved that AB \( \parallel \) DC, the quadrilateral ABCD fits the definition of a trapezium.
In simple words: If the two triangles formed by the intersecting diagonals of a quadrilateral, which are not sharing a common side, have equal areas, then adding the shared central triangle's area to each will show that the two larger triangles (formed by the main base and opposite vertices) have equal areas. This equality, combined with sharing a base, proves that the quadrilateral is a trapezium because their top vertices must be parallel to the shared base.

Exam Tip: This is the converse of a common area theorem. The key is to add a common area (like \( ar(\Delta AOB) \)) to both sides of the given area equality to obtain two larger triangles (\( \Delta ABD \) and \( \Delta ABC \)) with equal areas and a common base. This directly implies the parallelism required for a trapezium.

Question 16. Given ar(ADRC) = ar(ADPC) and ar(BDP) = ar(ARC), show that both the quadrilateral ABCD and DCPR are trapeziums.
Answer:
This question has two parts, showing two quadrilaterals are trapeziums based on given area equalities.

**Part 1: Prove that DCPR is a trapezium.**
Given: \( ar(\Delta DRC) = ar(\Delta DPC) \).
Consider \( \Delta DRC \) and \( \Delta DPC \). Both of these triangles share a common base DC. If two triangles have the same base and equal areas, then their vertices not on the common base must lie on a line parallel to the common base. In this case, the vertices not on DC are R and P.
Therefore, the line segment RP must be parallel to the line segment DC. That is, RP \( \parallel \) DC.
A quadrilateral is a trapezium if exactly one pair of its opposite sides is parallel.
Since we have shown that RP \( \parallel \) DC, the quadrilateral DCPR is a trapezium.

**Part 2: Prove that ABCD is a trapezium.**
Given: \( ar(\Delta BDP) = ar(\Delta ARC) \).
We already know from Part 1 that RP \( \parallel \) DC. (This is from \( ar(\Delta DRC) = ar(\Delta DPC) \)).
Let's consider the given equality \( ar(\Delta BDP) = ar(\Delta ARC) \).
We can write \( ar(\Delta BDP) = ar(\Delta BDC) + ar(\Delta DCP) \).
We can write \( ar(\Delta ARC) = ar(\Delta ADC) + ar(\Delta DRC) \).
So, \( ar(\Delta BDC) + ar(\Delta DCP) = ar(\Delta ADC) + ar(\Delta DRC) \).
From the first given condition, \( ar(\Delta DRC) = ar(\Delta DPC) \). Let's use this.
Substituting \( ar(\Delta DRC) \) for \( ar(\Delta DPC) \) (or vice versa):
\( ar(\Delta BDC) + ar(\Delta DRC) = ar(\Delta ADC) + ar(\Delta DRC) \)
Subtract \( ar(\Delta DRC) \) from both sides:
\( ar(\Delta BDC) = ar(\Delta ADC) \).

Now, consider \( \Delta BDC \) and \( \Delta ADC \). Both of these triangles share a common base DC. Since \( ar(\Delta BDC) = ar(\Delta ADC) \) and they share the base DC, their vertices not on the common base (B and A) must lie on a line parallel to the common base DC.
Therefore, the line segment AB must be parallel to the line segment DC. That is, AB \( \parallel \) DC.
A quadrilateral is a trapezium if exactly one pair of its opposite sides is parallel.
Since we have proved that AB \( \parallel \) DC, the quadrilateral ABCD is a trapezium.

Detailed steps:
**To prove DCPR is a trapezium:**
Given: \( ar(\Delta DRC) = ar(\Delta DPC) \) ..........(1)
Consider \( \Delta DRC \) and \( \Delta DPC \). They are on the same base DC.
Since their areas are equal, they must lie between the same parallel lines.
Therefore, RP \( \parallel \) DC.
Thus, DCPR is a trapezium (a quadrilateral with one pair of opposite sides parallel).

**To prove ABCD is a trapezium:**
Given: \( ar(\Delta BDP) = ar(\Delta ARC) \).
We can write \( ar(\Delta BDP) = ar(\Delta BDC) + ar(\Delta DPC) \).
We can write \( ar(\Delta ARC) = ar(\Delta ADC) + ar(\Delta DRC) \).
So, \( ar(\Delta BDC) + ar(\Delta DPC) = ar(\Delta ADC) + ar(\Delta DRC) \).
From (1), we know \( ar(\Delta DPC) = ar(\Delta DRC) \). Substitute this into the equation:
\( ar(\Delta BDC) + ar(\Delta DRC) = ar(\Delta ADC) + ar(\Delta DRC) \)
Subtract \( ar(\Delta DRC) \) from both sides:
\( ar(\Delta BDC) = ar(\Delta ADC) \).

Consider \( \Delta BDC \) and \( \Delta ADC \). They are on the same base DC.
Since their areas are equal, they must lie between the same parallel lines.
Therefore, AB \( \parallel \) DC.
Thus, ABCD is a trapezium (a quadrilateral with one pair of opposite sides parallel).
In simple words: If two triangles that share a base have the same area, it means their top points lie on a line that is parallel to their shared base. We apply this rule twice in this problem: first to show that one pair of sides is parallel for one quadrilateral, and then by manipulating the areas, we show another pair of sides is parallel for the second quadrilateral, proving both are trapeziums.

Exam Tip: This problem hinges on the converse of the area theorem: if two triangles on the same base have equal areas, they lie between the same parallel lines. The trick is to cleverly use the given area equalities to derive new ones that apply to common bases.