ML Aggarwal Class 9 Maths Solutions Chapter 10 Triangles

Question 1(a). In the figure given below, D, E and F are mid-points of the sides BC, CA and AB respectively of △ ABC. If AB = 6 cm, BC = 4.8 cm and CA = 5.6 cm, find the perimeter of (i) the trapezium FBCE (ii) the triangle DEF.
Answer:
(i) Because F is the midpoint of AB and E is the midpoint of AC, by the mid-point theorem, FE runs parallel to BC and measures half of BC. Therefore, FE = 2.4 cm. Similarly, FB = 3 cm and EC = 2.8 cm. Adding all four sides: FE + EC + CB + BF = 2.4 + 2.8 + 4.8 + 3 = 13 cm.

(ii) Using the mid-point theorem on triangle ABC with the three midpoints D, E, and F: FE equals 2.4 cm, FD equals 2.8 cm, and ED equals 3 cm. The sum of these three sides is 2.4 + 2.8 + 3 = 8.2 cm.
In simple words: When you connect the midpoints of a triangle's sides, each new side is half as long as the opposite side of the original triangle. Just add up all the sides to get the perimeter.

Exam Tip: Always apply the mid-point theorem carefully - the segment joining two midpoints is parallel to the third side and equals half its length. Verify your arithmetic when summing the perimeter.

Question 1(b). In the figure given below, D and E are midpoints of the sides AB and AC respectively. If BC = 5.6 cm and ∠B = 72°, compute (i) DE (ii) ∠ADE.
Answer:
(i) Since D and E are the midpoints of sides AB and AC, the mid-point theorem tells us that DE runs parallel to BC and its length is half of BC. Thus, DE = 2.8 cm.

(ii) Because DE is parallel to BC, corresponding angles formed by a transversal are equal. This means ∠ADE = ∠ABC = 72°.
In simple words: When a line inside a triangle connects two midpoints, it is half as long as the side it's parallel to. Parallel lines cut by a transversal make equal angles in matching positions.

Exam Tip: Remember that parallel lines create equal corresponding angles. This property often helps you find unknown angles without measuring them.

Question 1(c). In the figure given below, D and E are midpoints of AB, BC respectively and DF || BC. Prove that DBEF is a parallelogram. Calculate AC if AF = 2.6 cm.
Answer:
In triangle ABC, since D is the midpoint of AB and DF runs parallel to BC, by the converse of the mid-point theorem, F must be the midpoint of AC. Because F and E are midpoints of AC and BC respectively, EF runs parallel to AB, which means EF || DB. We are given that DF || BC, and since E lies on BC, we have DF || BE. By definition, a quadrilateral with both pairs of opposite sides parallel is a parallelogram, so DBEF is a parallelogram.

Since F is the midpoint of AC, the full length of AC is twice the length of AF. Therefore, AC = 2 × 2.6 = 5.2 cm.
In simple words: If a line through a midpoint runs parallel to a side, it hits the third side at its midpoint too. A shape where both opposite sides are parallel is called a parallelogram. A midpoint divides a line into two equal parts.

Exam Tip: The converse of the mid-point theorem is crucial here - it lets you identify F as a midpoint. Always show both pairs of opposite sides are parallel when proving a quadrilateral is a parallelogram.

Question 2. Prove that four triangles formed by joining in pairs, the mid-points of the sides of a triangle are congruent to each other.
Answer:
Let triangle ABC have midpoints D, E, and F on sides AB, BC, and CA respectively. Connect these midpoints to form segments DE, EF, and FD.

First, applying the mid-point theorem: DE is parallel to FC and DE equals half of AC (from triangle ABC). Also, DF is parallel to EC and DF equals half of BC. These conditions mean DECF is a parallelogram. A parallelogram's diagonal divides it into two congruent triangles, so △DEF ≅ △ECF.

Next, DE is parallel to AF and FE is parallel to AD, making ADEF a parallelogram. Its diagonal FD creates two congruent triangles: △DEF ≅ △AFD.

Finally, DF is parallel to BE and FE is parallel to DB, making DBEF a parallelogram. Its diagonal DE creates two congruent triangles: △DEF ≅ △DBE.

Combining all three results: △ADF ≅ △DBE ≅ △ECF ≅ △DEF.
In simple words: When you connect the three midpoints of a triangle, you create four smaller triangles that are all exactly the same size and shape as each other - they are congruent.

Exam Tip: Show the formation of each parallelogram clearly and identify the diagonal that divides it. This structured approach makes the proof easier to follow and ensures you don't miss any step.

Question 3. If D, E and F are mid-points of the sides AB, BC and CA respectively of an isosceles triangle, ABC, prove that △DEF is also isosceles.
Answer:
Given that triangle ABC is isosceles with AB = AC = x, and D, E, F are midpoints of AB, BC, CA respectively.

By the mid-point theorem applied to triangle ABC: DE is parallel to AC and DE = (1/2)AC = x/2.

Applying the mid-point theorem again: FE is parallel to AB and FE = (1/2)AB = x/2.

Since DE = FE = x/2, triangle DEF has two sides of equal length, making it isosceles.
In simple words: In an isosceles triangle, the two equal sides are the same length. When you connect the midpoints, two of the new sides turn out to be equal because they each measure half of one of the original equal sides.

Exam Tip: Always identify which two sides of the original triangle are equal, then apply the mid-point theorem to show that two sides of the midpoint triangle are also equal.

Question 4. The diagonals AC and BD of a parallelogram ABCD intersect at O. If P is the mid-point of AD, prove that (i) PO || AB (ii) PO = (1/2)CD.
Answer:
(i) In a parallelogram, the diagonals bisect each other, so O is the midpoint of BD. In triangle ABD, P and O are midpoints of sides AD and BD respectively. By the mid-point theorem, PO runs parallel to AB and equals half of AB.

(ii) Since ABCD is a parallelogram, opposite sides are equal: AB = CD. From part (i), PO = (1/2)AB. Combining these facts, PO = (1/2)CD.
In simple words: A line connecting the midpoints of two sides of a triangle is parallel to the third side and half its length. In a parallelogram, opposite sides are the same length.

Exam Tip: Recognize which triangle to use - triangle ABD is key here because O and P are midpoints of two of its sides. Do not confuse the properties of parallelograms with those of triangles.

Question 5. In the adjoining figure, ABCD is a quadrilateral in which P, Q, R and S are midpoints of AB, BC, CD and DA respectively. AC is its diagonal. Show that (i) SR || AC and SR = (1/2)AC (ii) PQ = SR (iii) PQRS is a parallelogram.
Answer:
(i) In triangle ADC, S and R are midpoints of sides AD and DC respectively. The mid-point theorem states that SR runs parallel to AC and equals half of AC.

(ii) In triangle ABC, P and Q are midpoints of sides AB and BC respectively. By the mid-point theorem, PQ runs parallel to AC and equals half of AC. Since both SR and PQ equal (1/2)AC, we have PQ = SR. Additionally, both are parallel to AC, so PQ || SR.

(iii) From part (ii), PQ = SR and PQ || SR. When a quadrilateral has one pair of opposite sides that are both equal and parallel, it is a parallelogram. Therefore, PQRS is a parallelogram.
In simple words: The midpoints of a quadrilateral's sides form a parallelogram. Each side of this midpoint shape is parallel to one of the quadrilateral's diagonals and measures half its length.

Exam Tip: Apply the mid-point theorem to different triangles formed by the diagonals. Showing that opposite sides are equal and parallel is the most straightforward way to prove a quadrilateral is a parallelogram.

Question 6. Show that the quadrilateral formed by joining the mid-points of the adjacent sides of a square, is also a square.
Answer:
Let ABCD be a square with E, F, G, H as midpoints of sides AB, BC, CD, DA respectively.

In triangle ACD, G and H are midpoints of CD and AD respectively, so GH || AC and GH = (1/2)AC.

In triangle ABC, E and F are midpoints of AB and BC respectively, so EF || AC and EF = (1/2)AC.

Therefore, EF || GH and EF = GH = (1/2)AC.

In triangle ABD, E and H are midpoints of AB and AD respectively, so EH || BD and EH = (1/2)BD.

In triangle BCD, F and G are midpoints of BC and CD respectively, so FG || BD and FG = (1/2)BD.

Therefore, EH || FG and EH = FG = (1/2)BD.

Since ABCD is a square, its diagonals AC and BD are equal in length and perpendicular. This means EF = GH = EH = FG, so all four sides of quadrilateral EFGH are equal. Additionally, since EF is parallel to AC and EH is parallel to BD, and AC ⊥ BD, we have EF ⊥ EH. All four angles of EFGH are therefore right angles. A quadrilateral with four equal sides and four right angles is a square, so EFGH is a square.
In simple words: When you connect the midpoints of a square's sides, you get another square that is rotated 45 degrees. All the new sides are equal, and all the angles are right angles.

Exam Tip: Use the fact that a square's diagonals are equal and perpendicular to each other. Show that all four sides of the midpoint quadrilateral are equal, and that adjacent sides are perpendicular.

Question 7. In the adjoining figure, AD and BE are medians of △ABC. If DF || BE, prove that CF = AC.
Answer: In △BCE, D is the midpoint of BC (since AD serves as a median). Since DF is parallel to BE, point F becomes the midpoint of CE (applying the converse of the mid-point theorem). This means CF equals half of CE, which can be written as CF = (1/2)CE. Given that BE is a median, we know that CE = (1/2)AC. Substituting this relationship into the previous equation gives us CF = (1/2) × (1/2)AC, which simplifies to CF = (1/4)AC.
In simple words: Since F is the midpoint of CE and BE is a median, CF works out to be one-quarter of the entire side AC.

Exam Tip: Always identify which segments are halves of others using the median and mid-point theorem properties - combining these relationships gives the final ratio.

Question 8. In the adjoining figure, ABCD is a parallelogram. E and F are mid-points of the sides AB and CD respectively. The straight lines AF and BF meet the straight lines ED and EC in points G and H respectively. Prove that (i) △HEB ≅ △HCF (ii) GEHF is a parallelogram.
Answer:
(i) Since ABCD forms a parallelogram, line FC runs parallel to line BE. Using properties of parallel lines with a transversal, we find that ∠CEB = ∠FCE (alternate angles), which means ∠HEB = ∠FCH. Similarly, ∠EBF = ∠CFB (alternate angles), so ∠EBH = ∠CFH. Because E and F mark the midpoints of AB and CD respectively, we have BE = (1/2)AB and CF = (1/2)CD. Since ABCD is a parallelogram, AB = CD. Dividing both sides by 2 gives (1/2)AB = (1/2)CD, so BE = CF. Now in △HEB and △HCF, we have shown that ∠HEB = ∠FCH, ∠EBH = ∠CFH, and BE = CF. By the ASA criterion for congruency, △HEB ≅ △HCF.(ii) From the fact that ABCD is a parallelogram, AB = CD. Since E and F are midpoints, AE = CF. Because AB runs parallel to CF (as AB lies on a line parallel to CD), we get AE || CF. Since AE = CF and AE || CF, quadrilateral AECF forms a parallelogram, which means AF || EC. From the figure, we observe that GF || EH. Similarly, AB = CD means DF = EB, and since DF || EB, quadrilateral DEBF becomes a parallelogram, giving us DE || FB. From the figure, GE || FH. Since GF || EH and GE || FH, quadrilateral GEHF is a parallelogram.
In simple words: The two triangles match using angle-side-angle, and GEHF forms a parallelogram because its opposite sides are parallel to each other.

Exam Tip: For congruency proofs, systematically list equal angles and sides in matching order - this approach prevents errors and ensures clarity for the examiner.

Question 9. ABC is an isosceles triangle with AB = AC. D, E and F are mid-points of the sides BC, AB and AC respectively. Prove that line segment AD is perpendicular to EF and is bisected by it.
Answer: In △ABD and △ACD, since △ABC is isosceles with AB = AC, the base angles are equal: ∠ABD = ∠ACD. Point D marks the midpoint of BC, so BD = CD. Combined with the given condition AB = AC, triangle ABD becomes congruent to triangle ACD by the SAS criterion. Therefore, ∠ADB = ∠ADC (by corresponding parts of congruent triangles). From the figure, angles ∠ADB and ∠ADC together form a straight line, so ∠ADB + ∠ADC = 180°. Since these angles are equal, 2∠ADB = 180°, giving ∠ADB = 90°. This establishes that AD is perpendicular to BC. Since D and E are midpoints of BC and AB respectively, the mid-point theorem tells us that DE || AC, or equivalently, DE || AF. Similarly, since D and F are midpoints of BC and AC respectively, DF || AB, or DF || AE. With both DE || AF and DF || AE, quadrilateral AEDF becomes a parallelogram. In any parallelogram, the diagonals bisect each other, so AD and EF bisect each other. Furthermore, because E and F are midpoints of AB and AC, the mid-point theorem ensures that EF || BC. Since AD is perpendicular to BC and EF runs parallel to BC, AD must be perpendicular to EF.
In simple words: AD is perpendicular to the base BC, and because EF is parallel to BC, AD is also perpendicular to EF. The diagonals of the parallelogram AEDF cut each other in half.

Exam Tip: Always verify perpendicularity by checking that angles sum to 180° - this step confirms the right angle and strengthens your proof structure.

Question 10(a). In the quadrilateral given below, AB || DC, E and F are mid-points of AD and BD respectively. Prove that (i) G is the mid-point of BC (ii) EG = (AB + DC).
Answer:
(i) In △ABD, since E is the midpoint of AD and F is the midpoint of BD, the mid-point theorem gives us EF || AB and EF = (1/2)AB. Given that AB || CD, and since EF || AB, we have EF || CD, which also means EG || CD. In △BCD, the fact that EG runs parallel to CD tells us that FG || CD. Since F is the midpoint of BD and FG || CD, by the converse of the mid-point theorem, G must be the midpoint of BC.(ii) In △BCD, F and G are midpoints of BD and BC respectively, so FG = (1/2)CD. Adding the equation EF = (1/2)AB from part (i) and the equation FG = (1/2)CD, we get: EF + FG = (1/2)AB + (1/2)CD EG = (1/2)(AB + CD)
In simple words: G divides BC exactly in half, and EG equals half the sum of the two parallel sides.

Exam Tip: Use the mid-point theorem and its converse repeatedly in trapezium problems - this systematic approach guarantees you capture all necessary relationships.

Question 10(b). In the quadrilateral given below, AB || DC || EG. If E is mid-point of AD, prove that (i) G is midpoint of BC (ii) 2EG = AB + CD
Answer:
(i) Since EG || AB, we can conclude that EF || AB. In △DAB, E is the midpoint of AD and EF is parallel to AB, so by the converse of the mid-point theorem, F becomes the midpoint of BD. This gives us EF = (1/2)AB. Since EG || DC, we have FG || DC. In △BCD, F is the midpoint of BD and FG is parallel to DC, so G must be the midpoint of BC by the converse of the mid-point theorem.(ii) In △BCD, F is the midpoint of BD and G is the midpoint of BC, giving us FG = (1/2)DC. Adding the equation EF = (1/2)AB and FG = (1/2)DC: EF + FG = (1/2)AB + (1/2)DC EG = (1/2)(AB + DC) 2EG = AB + CD
In simple words: When all three lines AB, EG, and DC are parallel, the segment EG is exactly halfway between them and equals half the sum of the outer parallel sides.

Exam Tip: The triple parallel condition (AB || DC || EG) restricts the geometry tightly - use this constraint to eliminate case work and apply theorems directly.

Question 10(c). In the quadrilateral given below, AB || DC. E and F are mid-points of non-parallel sides AD and BC respectively. Calculate: (i) EF if AB = 6 cm and DC = 4 cm (ii) AB if DC = 8 cm and EF = 9 cm.
Answer: ABCD is a trapezium where AB runs parallel to DC. E and F mark the midpoints of the non-parallel sides AD and BC respectively. To solve this, extend line CE to meet line BA extended at point G. In △EDC and △EAG, we have ED = EA (because E is the midpoint of AD), ∠CED = ∠GEA (vertically opposite angles), and ∠ECD = ∠EGA (alternate angles). By the ASA criterion, △EDC ≅ △EAG. From this congruence, DC = AG and EC = EG. Since F is the midpoint of BC, F becomes the midpoint of CG, which means EF passes through the midpoints and is parallel to both AB and DC. By the trapezium mid-segment theorem, EF = (1/2)(AB + DC).(i) Using the formula EF = (1/2)(AB + DC): EF = (1/2)(6 + 4) = (1/2)(10) = 5 cm(ii) Using EF = (1/2)(AB + DC) with DC = 8 cm and EF = 9 cm: 9 = (1/2)(AB + 8) 18 = AB + 8 AB = 10 cm
In simple words: The segment joining the midpoints of the non-parallel sides of a trapezium always equals half the sum of the two parallel sides.

Exam Tip: Always state the trapezium mid-segment formula at the start of your working - this shows the examiner you understand the core concept, even if you make an arithmetic error later.

Question 11. (i) If AB = 6 cm and DC = 4 cm, find EF. (ii) If DC = 8 cm and EF = 9 cm, find AB.
Answer: From the previous result, we have \( EF = \frac{1}{2}(AB + CD) \).

(i) Substituting AB = 6 cm and DC = 4 cm:
\( EF = \frac{1}{2}(6 + 4) = \frac{1}{2} \times 10 = 5 \text{ cm} \)
Thus, EF = 5 cm.

(ii) Substituting DC = 8 cm and EF = 9 cm:
\( 9 = \frac{1}{2}(AB + 8) \)
\( 18 = AB + 8 \)
\( AB = 10 \text{ cm} \)
Thus, AB = 10 cm.
In simple words: When you know three of these four measurements - the two parallel sides and the segment joining the midpoints - you can always find the fourth one using the formula.

Exam Tip: Always write the midpoint theorem formula first, then substitute the given values carefully to solve for the unknown.

Question 11(a). In the quadrilateral given below, AD = BC, P, Q, R and S are mid-points of AB, BD, CD and AC respectively. Prove that PQRS is a rhombus.
Answer: We are given that AD = BC. Consider triangle ABD where P and Q lie at the midpoints of AB and BD respectively. By the midpoint theorem, \( PQ \parallel AD \) and \( PQ = \frac{1}{2}AD = \frac{1}{2}BC \) ... (i)

In triangle ACD, R and S are the midpoints of CD and AC respectively. By the midpoint theorem, \( RS \parallel AD \) and \( RS = \frac{1}{2}AD = \frac{1}{2}BC \) ... (ii)

In triangle BCD, R and Q are the midpoints of CD and BD respectively. By the midpoint theorem, \( RQ \parallel BC \) and \( RQ = \frac{1}{2}BC \) ... (iii)

In triangle ABC, P and S are the midpoints of AB and AC respectively. By the midpoint theorem, \( PS \parallel BC \) and \( PS = \frac{1}{2}BC \) ... (iv)

From statements (i) and (ii): \( PQ \parallel RS \)
From statements (iii) and (iv): \( RQ \parallel PS \)

From all four statements, we see that \( PQ = RS = PS = RQ \).

Since all four sides are equal in length and opposite sides are parallel, PQRS is a rhombus.
In simple words: When you join the midpoints of a quadrilateral's sides, if two opposite sides of the original quadrilateral are equal, the shape you get has all four sides equal, making it a rhombus.

Exam Tip: The key is applying the midpoint theorem to four different triangles and showing that all four segments connecting the midpoints are equal.

Question 11(b). In the figure given below, ABCD is a kite in which BC = CD, AB = AD. E, F, G are mid-points of CD, BC and AB respectively. Prove that: (i) ∠EFG = 90° (ii) The line drawn through G and parallel to FE bisects DA.
Answer:

Construction: Join AC and BD; let them intersect at O. Join EF and FG.

Part (i): Proving ∠EFG = 90°

A key property of kites is that the diagonals intersect at right angles. Therefore, \( \angle MON = 90° \) ... (i)

In triangle BCD, E and F are the midpoints of CD and BC respectively. By the midpoint theorem, \( EF \parallel DB \) and \( EF = \frac{1}{2}DB \) ... (ii)

Since \( EF \parallel DB \), we can say that \( MF \parallel ON \).

When two lines are parallel in a quadrilateral (MFNO), the sum of opposite angles equals 180°. Therefore,
\( \angle MON + \angle MFN = 180° \)
\( 90° + \angle MFN = 180° \)
\( \angle MFN = 90° \)

From the figure, \( \angle EFG = \angle MFN = 90° \), so the angle is proved.

Part (ii): The line through G parallel to FE bisects DA

From part (i), we established that \( FE \parallel BD \).

Let the line through G that is parallel to FE be called GH. Then \( GH \parallel FE \), which means \( GH \parallel BD \).

In triangle ABD, GH is parallel to BD, and G is the midpoint of AB. By the converse of the midpoint theorem, H must be the midpoint of AD.

Therefore, the line drawn through G and parallel to FE bisects DA.
In simple words: The angles of a kite's diagonals are always 90 degrees. When you join the midpoints of a kite's sides in a certain way, you create a right angle. A line through one midpoint that is parallel to another creates a path that cuts the opposite side exactly in half.

Exam Tip: Remember that kites have perpendicular diagonals - this is the crucial fact. For part (ii), use the converse of the midpoint theorem, not the theorem itself.

Question 12. In the adjoining figure, the lines l, m and n are parallel to each other, and G is mid-point of CD. Calculate: (i) BG if AD = 6 cm (ii) CF if GE = 2.3 cm (iii) AB if BC = 2.4 cm (iv) ED if FD = 4.4 cm
Answer:

(i) In triangle ACD, G is the midpoint of CD and BG runs parallel to AD. By the converse of the midpoint theorem, B is the midpoint of AC. Using the midpoint theorem,
\( BG = \frac{1}{2}AD = \frac{1}{2} \times 6 = 3 \text{ cm} \)
Thus, BG = 3 cm.

(ii) In triangle CDF, G is the midpoint of CD and GE runs parallel to CF. By the converse of the midpoint theorem, E is the midpoint of FD. Using the midpoint theorem,
\( GE = \frac{1}{2}CF \)
\( CF = 2 \times GE = 2(2.3) = 4.6 \text{ cm} \)
Thus, CF = 4.6 cm.

(iii) From part (i), we found that B is the midpoint of AC. This means:
\( AB = BC = 2.4 \text{ cm} \)
Thus, AB = 2.4 cm.

(iv) From part (ii), we found that E is the midpoint of FD. This means:
\( ED = \frac{1}{2}FD = \frac{1}{2} \times 4.4 = 2.2 \text{ cm} \)
Thus, ED = 2.2 cm.
In simple words: When parallel lines cut two sides of a triangle, the midpoint relationships help you find unknown lengths by using the fact that a midpoint divides a side into two equal parts.

Exam Tip: In each part, first identify which triangle you are working with and which sides are parallel. Then apply either the midpoint theorem or its converse as needed.

Question 1. In a △ABC, AB = 3 cm, BC = 4 cm and CA = 5 cm. If D and E are mid-points of AB and BC respectively, then the length of DE is
(a) 1.5 cm
(b) 2 cm
(c) 2.5 cm
(d) 3.5 cm
Answer: (c) 2.5 cm
In simple words: When you connect the midpoints of two sides of a triangle, the segment you draw is parallel to the third side and measures exactly half its length.

Exam Tip: This is a direct application of the midpoint theorem - always check if the given points are midpoints of two sides, and if so, the connecting segment is half the third side.

Question 2. In the adjoining figure, ABCD is a rectangle in which AB = 6 cm and AD = 8 cm. If P and Q are mid-points of the sides BC and CD respectively, then the length of PQ is
(a) 7 cm
(b) 5 cm
(c) 4 cm
(d) 3 cm
Answer: (b) 5 cm
In simple words: Start by finding the diagonal BD using the Pythagorean theorem. Since P and Q are midpoints of two sides that meet at one corner, the segment PQ is half the length of the diagonal opposite to that corner.

Exam Tip: In rectangle problems with midpoints, always compute the relevant diagonal first using \( \sqrt{a^2 + b^2} \), then apply the midpoint theorem to that diagonal within the appropriate triangle.

Question 3. D and E are mid-points of the sides AB and AC of △ABC and O is any point on the side BC. O is joined to A. If P and Q are mid-points of OB and OC respectively, then DEQP is
(a) a square
(b) a rectangle
(c) a rhombus
(d) a parallelogram
Answer: (d) a parallelogram
In simple words: When you create a quadrilateral by joining four midpoints (one pair from two triangles that share a common vertex), opposite sides of that quadrilateral will always be parallel to each other, making it a parallelogram.

Exam Tip: Apply the midpoint theorem separately to three different triangles (ABC, ABO, and ACO) to show that opposite sides are parallel, which proves the quadrilateral is a parallelogram.

Question 4. The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle if
(a) PQRS is a parallelogram
(b) PQRS is a rectangle
(c) the diagonals of PQRS are perpendicular to each other
(d) the diagonals of PQRS are equal.
Answer: (c) the diagonals of PQRS are perpendicular to each other
In simple words: When you join the midpoints of any quadrilateral, you always get a parallelogram. It becomes a rectangle only when the original quadrilateral's two diagonals cross at right angles.

Exam Tip: The midpoint quadrilateral is always a parallelogram. Its shape changes (becoming a rectangle, rhombus, or square) based only on the angle between the original quadrilateral's diagonals - perpendicular diagonals give you a rectangle.

Question 5. The quadrilateral formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order is a rhombus if
(1) ABCD is a parallelogram
(2) ABCD is a rhombus
(3) the diagonals of ABCD are equal
(4) the diagonals of ABCD are perpendicular to each other.
Answer: Let ABCD be a quadrilateral with P, Q, R and S as midpoints of AB, BC, CD and DA respectively. Let the diagonals have equal length, meaning AC = BD = x.

In \( \triangle BCA \), since P and Q are midpoints of AB and BC respectively:
\[ PQ \parallel AC \text{ and } PQ = \frac{1}{2}AC = \frac{1}{2}x \text{ (By midpoint theorem)} \text{ ........(1)} \]

In \( \triangle ACD \), since S and R are midpoints of AD and CD respectively:
\[ SR \parallel AC \text{ and } SR = \frac{1}{2}AC = \frac{1}{2}x \text{ (By midpoint theorem)} \text{ ........(2)} \]

In \( \triangle ABD \), since S and P are midpoints of AD and AB respectively:
\[ SP \parallel BD \text{ and } SP = \frac{1}{2}BD = \frac{1}{2}x \text{ (By midpoint theorem) ........(3)} \]

In \( \triangle BCD \), since Q and R are midpoints of BC and CD respectively:
\[ QR \parallel BD \text{ and } QR = \frac{1}{2}BD = \frac{1}{2}x \text{ (By midpoint theorem) ........(4)} \]

From statements (1), (2), (3) and (4): \( PQ = SR = SP = QR \). Therefore, PQRS is a rhombus. The quadrilateral formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order is a rhombus if the diagonals of ABCD are equal.
In simple words: When a quadrilateral's two diagonals are the same length, connecting the midpoints of its four sides creates a rhombus shape with all four sides equal.

Exam Tip: Always apply the midpoint theorem to each of the four triangles formed by the diagonals, and match each side of PQRS to half a diagonal to show all sides are equal.

Question 6. The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if
(1) ABCD is a rhombus
(2) diagonals of ABCD are equal
(3) diagonals of ABCD are perpendicular to each other
(4) diagonals of ABCD are equal and perpendicular to each other.
Answer: Let ABCD be a quadrilateral with P, Q, R and S as midpoints of AB, BC, CD and DA respectively. Let the diagonals satisfy AC = BD = x and AC \( \perp \) BD.

In \( \triangle BCA \), since P and Q are midpoints of AB and BC respectively:
\[ PQ \parallel AC \text{ and } PQ = \frac{1}{2}AC = \frac{1}{2}x \text{ (By midpoint theorem) ........(1)} \]

In \( \triangle ACD \), since S and R are midpoints of AD and CD respectively:
\[ SR \parallel AC \text{ and } SR = \frac{1}{2}AC = \frac{1}{2}x \text{ (By midpoint theorem) ........(2)} \]

In \( \triangle ABD \), since S and P are midpoints of AD and AB respectively:
\[ SP \parallel BD \text{ and } SP = \frac{1}{2}BD = \frac{1}{2}x \text{ (By midpoint theorem) ........(3)} \]

In \( \triangle BCD \), since Q and R are midpoints of BC and CD respectively:
\[ QR \parallel BD \text{ and } QR = \frac{1}{2}BD = \frac{1}{2}x \text{ (By midpoint theorem) ........(4)} \]

From (1), (2), (3) and (4): \( PQ = SR = SP = QR \). Thus PQRS is a rhombus.

Since \( SP \parallel BD \) and AC \( \perp \) BD, we have \( SP \perp AC \). Since \( SR \parallel AC \) and AC \( \perp \) BD, we have \( SR \perp BD \).

In quadrilateral OMSN, vertically opposite angles are equal: \( \angle OMS = \angle DMR \). The sum of angles in a quadrilateral is 360°, so \( \angle O + \angle M + \angle N + \angle S = 360° \). Since three angles are 90° each: \( 90° + 90° + 90° + \angle S = 360° \), giving \( \angle S = 90° \).

In a rhombus, adjacent angles sum to 180°. From \( \angle S + \angle R = 180° \): \( \angle R = 90° \). From \( \angle Q + \angle R = 180° \): \( \angle Q = 90° \). From \( \angle S + \angle P = 180° \): \( \angle P = 90° \).

Since \( PQ = QR = RS = SP \) and all angles equal 90°, PQRS is a square. The figure formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a square only if the diagonals of ABCD are equal and perpendicular to each other.
In simple words: For the midpoint quadrilateral to be a square, the original quadrilateral must have diagonals that are both the same length and meet at a right angle.

Exam Tip: Show that PQRS is first a rhombus (all sides equal), then prove all angles are 90° using the perpendicularity and angle sum properties - these two conditions together define a square.

Question 7. Consider the following two statements:
Statement 1: The line segment joining the midpoints of any two sides of a triangle is parallel to the third side and is equal to half of it.
Statement 2: The line through the midpoint of one side of a triangle and parallel to another side bisects the third side.
Which of the following is valid?
(1) Both the statements are true.
(2) Both the statements are false.
(3) Statement 1 is true, and Statement 2 is false.
(4) Statement 1 is false, and Statement 2 is true.
Answer: Statement 1 describes the midpoint theorem: the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and is equal to half of it. This statement is true.

Statement 2 describes the converse of the midpoint theorem: the line through the midpoint of one side of a triangle and parallel to another side bisects the third side. This statement is also true.
In simple words: Both statements are correct. The first says connecting two midpoints gives a line parallel to the third side and half its length. The second says if a line starts at a midpoint and runs parallel to another side, it will cut the third side in half.

Exam Tip: These are the midpoint theorem and its converse - both are standard results. Always recall that "midpoint + parallel = bisects" (the converse) to distinguish it from the forward theorem.

Question. Assertion (A): For a \( \triangle ABC \), line segment EF is drawn such that E is the midpoint of AB and F is a midpoint of AC. Then the quadrilateral formed EFCB is a trapezium.
Reason (R): The line segment joining the midpoint of two sides of a triangle is parallel to the third side.
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: The Midpoint Theorem states: "The line segment joining the midpoints of any two sides of a triangle is parallel to the third side and is equal to half of its length." Therefore, Reason (R) is true.

In \( \triangle ABC \): E is the midpoint of AB and F is the midpoint of AC. The line segment joining E and F is EF. By the Midpoint Theorem, EF is parallel to BC. The quadrilateral EFCB has one pair of parallel sides (EF and BC). A quadrilateral with exactly one pair of parallel sides is defined as a trapezium. Therefore, EFCB is indeed a trapezium, so Assertion (A) is true.

Since both Assertion (A) and Reason (R) are true, and Reason (R) directly explains why EFCB is a trapezium (because EF is parallel to BC by the Midpoint Theorem), Reason (R) is the correct reason for Assertion (A).
In simple words: E and F are midpoints, so segment EF is parallel to BC. This makes EFCB a trapezium because it has one pair of parallel sides.

Exam Tip: A trapezium requires exactly one pair of parallel sides. Use the Midpoint Theorem to establish the parallel sides, then apply the definition of a trapezium to confirm the assertion.

Question. Assertion (A): In a \( \triangle DEF \), we have DE = EF = DF = 6 cm. A line segment PQ is drawn parallel to DF such that EP = 3 cm. Then we can conclude that PQ = 3 cm.
Reason (R): Any line segment drawn inside a triangle parallel to the base of the triangle cuts the remaining two sides in half.
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: According to the converse of the midpoint theorem, a straight line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side.

It is given that PQ is drawn parallel to DF and EP = 3 cm. Since DE = 6 cm and EP = 3 cm, point P is the midpoint of ED. By the converse of the midpoint theorem, Q will also be the midpoint of EF.

The Midpoint Theorem states that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and is equal to half of it. Therefore:
\[ PQ = \frac{1}{2} \times DF = \frac{1}{2} \times 6 \text{ cm} = 3 \text{ cm} \]
Thus, Assertion (A) is true.

Reason (R) states that any line segment parallel to the base cuts the other two sides in half. However, this is not always true - a line parallel to the base cuts the other two sides in half only if it passes through the midpoints. A line segment parallel to the base does not necessarily cut the other two sides in half unless it passes through the midpoints. Therefore, Reason (R) is false.
In simple words: Since EP is half of DE (both 3 and 6 cm), point P is a midpoint. By the midpoint theorem, PQ must be half of DF, which is 3 cm. However, the reason given is wrong because parallel lines don't always bisect sides - they must pass through midpoints.

Exam Tip: Distinguish between "any line parallel to a side" (which may not bisect) and "a line through a midpoint parallel to a side" (which always bisects). Use the converse of the midpoint theorem when given parallel lines and one known distance.

Question. Assertion (A): Refer to the adjoining figure. Three lines p, q, r are parallel to each other and PQ = QR = 1 cm. Then we conclude that \( AB = \frac{1}{2}AC \).
Reason (R): If a transversal makes equal intercepts on three parallel lines, then another transversal will also make equal intercepts.
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: According to the equal intercept theorem, if a transversal makes equal intercepts on three or more parallel lines, then any other line cutting them will also make equal intercepts.

Given: Three lines p, q, r are parallel to each other, with PQ = QR = 1 cm. This means the transversal on the right (line \( l \)) makes equal intercepts on the three parallel lines. By the equal intercept theorem, any other transversal cutting these same three parallel lines will also make equal intercepts. Therefore, on transversal m, we have AB = BC, which means \( AB = \frac{1}{2}AC \). Thus, Assertion (A) is true.

Reason (R) correctly states the equal intercept theorem. This theorem directly justifies the conclusion that if one transversal makes equal intercepts on three parallel lines, another transversal will also make equal intercepts. Therefore, Reason (R) is true and is the correct reason for Assertion (A).
In simple words: When one line cutting three parallel lines creates equal spaces, any other line cutting those same parallels will also create equal spaces. Since PQ = QR on one transversal, AB = BC on another transversal, making AB half of AC.

Exam Tip: Apply the equal intercept theorem directly - equal spacing on one transversal guarantees equal spacing on any other transversal cutting the same parallel lines. This is a powerful tool for finding lengths.

Question. Assertion (A): If three parallel lines cut off equal segments on one transversal, then they cut off equal segments on any other transversal also. Reason (R): By the equal intercept theorem, if three parallel lines intercept equal distances on a transversal, they will also intercept equal distances on any transversal crossing them.
Answer: Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A). When three lines p, q, r are parallel with PQ = QR = 1 cm on one transversal, the equal intercept theorem tells us that AB = BC on any other transversal. From the figure, since AC = AB + BC and AB = BC, we can write AC = 2AB, which means AB = (1/2)AC. This proves the assertion is correct and supported by the reason given.
In simple words: When three lines run parallel to each other and make equal parts on one line crossing them, they will make equal parts on any other line crossing them too. This rule always works.

Exam Tip: For assertion-reason questions, verify both statements separately, then check if the reason actually explains the assertion. Both being true is not enough - the reason must logically support the assertion for full marks.

Chapter Test

Question 1. ABCD is a rhombus with P, Q and R as midpoints of AB, BC and CD respectively. Prove that PQ ⊥ QR.
Answer: Connect AC and BD. The diagonals of a rhombus meet at right angles, so ∠MON = 90°. In triangle BCD, since Q and R are midpoints of BC and CD, the line segment RQ is parallel to diagonal DB and equals half its length. Since RQ is parallel to DB, this means MQ is parallel to ON. Using the property that co-interior angles sum to 180°, we have ∠MON + ∠MQN = 180°. Since ∠MON = 90°, we get ∠MQN = 90°. Therefore, PQ and QR are perpendicular to each other.
In simple words: The diagonals of a rhombus cross at a 90-degree angle. Using the midpoint theorem, we can show that the angle at Q must also be 90 degrees, making PQ and QR perpendicular.

Exam Tip: Use the property that diagonals of a rhombus are perpendicular. Apply the midpoint theorem to connect segment lengths with parallel lines. The co-interior angle property is key to finding the perpendicular angle.

Question 2. The diagonals of a quadrilateral ABCD are perpendicular. Show that the quadrilateral formed by joining the mid-points of its adjacent sides is a rectangle.
Answer: Consider quadrilateral ABCD with perpendicular diagonals AC and BD. Let P, Q, R, S be the midpoints of sides AB, BC, CD, and DA respectively. In triangle ABC, since P and Q are midpoints of AB and BC, line segment PQ is parallel to AC and equals half its length - this is the midpoint theorem. Similarly, in triangle ADC, since S and R are midpoints of AD and DC, line segment SR is parallel to AC and equals half its length. Therefore, PQ is parallel to SR and PQ = SR, making PQRS a parallelogram. Next, in triangle ABD, since S and P are midpoints of AD and AB, line segment SP is parallel to BD and equals half its length. Since AC and BD are perpendicular, SP must be perpendicular to AC. From our earlier work, SR is parallel to AC, so SP is perpendicular to SR. This means ∠RSP = 90°. Since PQRS is a parallelogram with one right angle, it must be a rectangle.
In simple words: When you connect the midpoints of a quadrilateral's sides, you always get a parallelogram. If the original quadrilateral's diagonals cross at right angles, that parallelogram becomes a rectangle with all corners at 90 degrees.

Exam Tip: Apply the midpoint theorem three times to establish parallel sides and equal lengths. The key is showing that the angle at one corner is 90 degrees using the perpendicular diagonals condition. Once you have a parallelogram with one right angle, it is a rectangle.

Question 3. If D, E and F are mid-points of the sides BC, CA and AB respectively of a △ABC, prove that AD and FE bisect each other.
Answer: Consider triangle ABC with D, E, F as midpoints of sides BC, CA, AB respectively. Since D and E are midpoints of BC and CA, by the midpoint theorem, segment DE is parallel to AB and equals half of AB. Since F is the midpoint of AB, we have AF = (1/2)AB. From this, AF = DE. Also, DE is parallel to AF. Similarly, since F and D are midpoints of AB and BC, segment FD is parallel to AC and equals half of AC. Since E is the midpoint of AC, we have AE = (1/2)AC. Therefore, FD = AE, and FD is parallel to AE. Combining all these results - DE is parallel to AF with DE = AF, and FD is parallel to AE with FD = AE - we conclude that AEDF is a parallelogram. In a parallelogram, the diagonals bisect each other. The diagonals of parallelogram AEDF are AD and EF, so they bisect each other.
In simple words: By the midpoint theorem, the segments connecting midpoints have special relationships. When you use these relationships, you can show that the four points A, E, D, F form a parallelogram. The diagonals of any parallelogram cut each other in half.

Exam Tip: Use the midpoint theorem multiple times to establish segment lengths and parallelism. Show that AEDF is a parallelogram by proving two pairs of opposite sides are equal and parallel. Then use the diagonal property of parallelograms.

Question 4. In △ABC, D and E are mid-points of the sides AB and AC respectively. Through E, a straight line is drawn parallel to AB to meet BC at F. Prove that BDEF is a parallelogram. If AB = 8 cm and BC = 9 cm, find the perimeter of the parallelogram BDEF.
Answer: Since D and E are midpoints of AB and AC respectively, by the midpoint theorem, DE is parallel to BC and DE = (1/2)BC. This can also be written as DE is parallel to BF and DE = (1/2)BC. Given that through point E a line is drawn parallel to AB meeting BC at F, by the converse of the midpoint theorem, F must be the midpoint of BC. Therefore, BF = (1/2)BC. From the above, we have DE = BF and DE is parallel to BF. Since F and E are midpoints of BC and AC respectively, by the midpoint theorem, FE is parallel to AB and FE = (1/2)AB. Since D is the midpoint of AB, we have BD = (1/2)AB. From this, FE = BD and FE is parallel to BD. We have shown that DE = BF with DE parallel to BF, and BD = FE with BD parallel to FE. Therefore, BDEF is a parallelogram. To find the perimeter: BD = (1/2)AB = (1/2)(8) = 4 cm. FE = (1/2)BC = (1/2)(9) = 4.5 cm. Perimeter of BDEF = 2(BD + FE) = 2(4 + 4.5) = 2(8.5) = 17 cm.
In simple words: The midpoint theorem helps us find segments that are parallel and equal. When opposite sides are both parallel and equal, the figure is a parallelogram. To find the perimeter, we use half the given lengths since the sides of the parallelogram are half the triangle's sides.

Exam Tip: Apply the midpoint theorem twice to establish that opposite sides are parallel and equal. For the perimeter calculation, remember that the sides of the parallelogram equal half the corresponding triangle sides. Show all four properties (two pairs of opposite sides) to prove it is a parallelogram.

Question 5. In the adjoining figure, ABCD is a parallelogram and E is mid-point of AD. DL || EB meets AB produced at F. Prove that B is mid-point of AF and EB = LF.
Answer: Given that DL is parallel to EB, we can conclude that BE is parallel to DF. Consider triangle AFD. Since E is the midpoint of AD and BE is parallel to DF, by the converse of the midpoint theorem, B must be the midpoint of AF. In quadrilateral BEDL, we have LD parallel to BE and BL parallel to DE, which means BEDL is a parallelogram. In a parallelogram, opposite sides are equal, so LD = BE. Let's say LD = BE = x. Since E is the midpoint of AD and B is the midpoint of AF, by the midpoint theorem applied to triangle AFD, BE equals half of FD. Therefore, FD = 2BE = 2x. Now, LF = FD - LD = 2x - x = x. Since LF = x and BE = x, we have shown that EB = LF.
In simple words: When a line through E is parallel to EB, it creates a parallelogram BEDL. Using the midpoint theorem and properties of parallelograms, we can show that B divides AF equally and that two segments have the same length.

Exam Tip: Recognize that parallel lines and midpoints combine to create a parallelogram. Use the converse of the midpoint theorem to locate B as a midpoint. Set up an algebraic expression for the segments to compare their lengths clearly.

Question 6. In the adjoining figure, ABCD is a parallelogram. If P and Q are mid-points of sides CD and BC respectively. Show that CR = (1/4)AC.
Answer: In parallelogram ABCD, the diagonals bisect each other. Therefore, AO = OC = (1/2)AC. In triangle BCD, since P and Q are midpoints of CD and BC respectively, by the midpoint theorem, PQ is parallel to BD. Since PQ is parallel to BD, this means QR is parallel to BO. Consider triangle BCO. Q is the midpoint of BC and QR is parallel to BO. By the converse of the midpoint theorem, R must be the midpoint of OC. Since R is the midpoint of OC, we have CR = (1/2)OC. Substituting OC = (1/2)AC from our earlier result, CR = (1/2) × (1/2)AC = (1/4)AC. Therefore, CR = (1/4)AC.
In simple words: The diagonals of a parallelogram split each other in half. Using the midpoint theorem twice - once in triangle BCD and once in triangle BCO - we can show that CR is one quarter of the full diagonal AC.

Exam Tip: Apply the midpoint theorem in two different triangles to progressively narrow down the length. The key is recognizing that diagonals of a parallelogram bisect each other, giving you the starting relationship. Then use midpoint properties twice to reach the final (1/4) result.