GSEB Class 9 Maths Solutions Chapter 8 Quadrilaterals Exercise 8.1

Question 1. The angles of a quadrilateral are in the ratio \( 3:5:9:13 \). Find all the angles of the quadrilateral.
Answer: We have angles of a quadrilateral in the ratio \( \angle A : \angle B : \angle C : \angle D = 3 : 5 : 9 : 13 \).
Let \( \angle A = 3x \),
\( \angle B = 5x \),
\( \angle C = 9x \),
and \( \angle D = 13x \).
The sum of all angles in a quadrilateral is \( 360^\circ \).
So, \( \angle A + \angle B + \angle C + \angle D = 360^\circ \).
\( 3x + 5x + 9x + 13x = 360^\circ \)
\( \implies 30x = 360^\circ \)
\( x = \frac{360^\circ}{30} \)
\( \implies x = 12^\circ \).
Hence, the angles are:
\( \angle A = 3x = 3 \times 12^\circ = 36^\circ \)
\( \angle B = 5x = 5 \times 12^\circ = 60^\circ \)
\( \angle C = 9x = 9 \times 12^\circ = 108^\circ \)
\( \angle D = 13x = 13 \times 12^\circ = 156^\circ \).
In simple words: The angles of a four-sided shape are in a given proportion. We find a common multiplier for this proportion and then calculate each angle, knowing that all angles in a quadrilateral add up to 360 degrees.

Exam Tip: Remember that the sum of the interior angles of any quadrilateral is always \( 360^\circ \). Use this fact to set up an equation when angles are given in a ratio.

Question 2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Answer:

Given: ABCD is a parallelogram where \( AC = BD \).
To prove: ABCD is a rectangle.
Proof: Consider \( \triangle ABC \) and \( \triangle BAD \).
1. \( AB = BA \) (Common side)
2. \( BC = AD \) (Opposite sides of a parallelogram)
3. \( AC = BD \) (Given)
Therefore, \( \triangle ABC \cong \triangle BAD \) (by SSS congruence rule).
This implies \( \angle ABC = \angle BAD \) (by CPCT - Corresponding Parts of Congruent Triangles).
Also, for a parallelogram, consecutive interior angles are supplementary.
So, \( \angle ABC + \angle BAD = 180^\circ \).
Since \( \angle ABC = \angle BAD \), we can write \( \angle BAD + \angle BAD = 180^\circ \).
\( \implies 2\angle BAD = 180^\circ \)
\( \angle BAD = \frac{180^\circ}{2} \)
\( \angle BAD = 90^\circ \).
A parallelogram with one angle equal to \( 90^\circ \) is a rectangle.
Therefore, ABCD is a rectangle.
In simple words: If a parallelogram has diagonals that are the same length, then it is a rectangle. We prove this by showing two triangles within the parallelogram are identical, which leads to one of its angles being 90 degrees.

Exam Tip: To prove a parallelogram is a rectangle, you need to show either that one of its angles is \( 90^\circ \) or that its diagonals are equal. This question uses the latter to prove the former.

Question 3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Answer:

Given: ABCD is a quadrilateral where diagonals AC and BD cut each other at right angles, meaning \( \angle AOB = 90^\circ \), and they bisect each other, so \( OA = OC \) and \( OB = OD \).
To prove: ABCD is a rhombus.
Proof: Consider \( \triangle AOB \) and \( \triangle AOD \).
1. \( OB = OD \) (Given that diagonals bisect each other)
2. \( \angle AOB = \angle AOD \) (Both are \( 90^\circ \) because diagonals bisect at right angles)
3. \( OA = OA \) (Common side)
Therefore, \( \triangle AOB \cong \triangle AOD \) (by SAS congruence rule).
This implies \( AB = AD \) (by CPCT).
Similarly, we can prove the following:
- \( \triangle AOB \cong \triangle COB \implies AB = CB \)
- \( \triangle COB \cong \triangle COD \implies CB = CD \)
- \( \triangle AOD \cong \triangle COD \implies AD = CD \)
From these results, we get \( AB = BC = CD = AD \).
Since all four sides of the quadrilateral are equal, ABCD is a rhombus.
In simple words: If a four-sided shape has diagonals that cut each other in half and meet at a perfect 90-degree angle, then all its sides must be equal. This means the shape is a rhombus.

Exam Tip: A rhombus is defined as a quadrilateral with all four sides equal. To prove a quadrilateral is a rhombus, show that its diagonals bisect each other at right angles or that all its sides are equal.

Question 4. Show that the diagonals of a square are equal and bisect each other at right angles.
Answer:

Given: ABCD is a square where AC and BD are its diagonals.
To Prove: (i) \( AC = BD \)
(ii) AC and BD bisect each other at right angles.
Proof:
**(i) Diagonals are equal (AC = BD):**
Consider \( \triangle ABC \) and \( \triangle BAD \).
1. \( BC = AD \) (Opposite sides of a square are equal)
2. \( \angle ABC = \angle BAD \) (Each angle of a square is \( 90^\circ \))
3. \( AB = BA \) (Common side)
Therefore, \( \triangle ABC \cong \triangle BAD \) (by SAS congruence rule).
This implies \( AC = BD \) (by CPCT). So, the diagonals are equal.

**(ii) Diagonals bisect each other at right angles:**
*To show they bisect each other (OA = OC and OB = OD):*
Consider \( \triangle AOB \) and \( \triangle COD \).
1. \( AB = CD \) (Opposite sides of a square are equal)
2. \( \angle OAB = \angle OCD \) (Alternate interior angles, since \( AB || CD \))
3. \( \angle OBA = \angle ODC \) (Alternate interior angles, since \( AB || CD \))
Therefore, \( \triangle AOB \cong \triangle COD \) (by ASA congruence rule).
This implies \( OA = OC \) and \( OB = OD \) (by CPCT). So, the diagonals bisect each other.

*To show they bisect at right angles (\( \angle AOB = 90^\circ \)):*
Consider \( \triangle AOD \) and \( \triangle COD \).
1. \( AD = CD \) (Sides of a square are equal)
2. \( OD = OD \) (Common side)
3. \( OA = OC \) (Proved above that diagonals bisect)
Therefore, \( \triangle AOD \cong \triangle COD \) (by SSS congruence rule).
This implies \( \angle AOD = \angle COD \) (by CPCT).
Since \( AC \) is a straight line, \( \angle AOD + \angle COD = 180^\circ \) (Linear pair axiom).
As \( \angle AOD = \angle COD \), we have \( \angle AOD + \angle AOD = 180^\circ \).
\( \implies 2\angle AOD = 180^\circ \)
\( \angle AOD = 90^\circ \).
Similarly, \( \angle AOB = \angle BOC = \angle COD = \angle DOA = 90^\circ \).
Thus, the diagonals of a square are equal and bisect each other at right angles.
In simple words: For a square, its two main lines that go from corner to corner are the same length, and they cut each other exactly in half. Where these lines cross, they form perfect right angles, meaning they are perpendicular.

Exam Tip: Remember the key properties of a square: all sides are equal, all angles are \( 90^\circ \), and its diagonals are equal, bisect each other, and are perpendicular.

Question 5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Answer:

Given: In quadrilateral ABCD, \( AC = BD \), \( OA = OC \), \( OB = OD \), and \( \angle AOB = 90^\circ \).
To Prove: ABCD is a square.
Proof:
*First, show ABCD is a parallelogram:*
Since diagonals bisect each other ( \( OA = OC \) and \( OB = OD \) ), ABCD is a parallelogram.

*Next, show ABCD is a rhombus (all sides equal):*
Consider \( \triangle AOB \) and \( \triangle COB \).
1. \( OA = OC \) (Given)
2. \( \angle AOB = \angle COB \) (Both are \( 90^\circ \) because \( \angle AOB = 90^\circ \) and \( \angle AOB + \angle COB = 180^\circ \), so \( \angle COB = 90^\circ \))
3. \( OB = OB \) (Common side)
Therefore, \( \triangle AOB \cong \triangle COB \) (by SAS congruence rule).
This implies \( AB = BC \) (by CPCT).
Since ABCD is a parallelogram with adjacent sides equal (\( AB = BC \)), it is a rhombus.
So, \( AB = BC = CD = DA \).

*Finally, show ABCD is a rectangle (all angles \( 90^\circ \)):*
Consider \( \triangle ABC \) and \( \triangle BAD \).
1. \( AB = BA \) (Common side)
2. \( BC = AD \) (Opposite sides of a rhombus are equal)
3. \( AC = BD \) (Given)
Therefore, \( \triangle ABC \cong \triangle BAD \) (by SSS congruence rule).
This implies \( \angle ABC = \angle BAD \) (by CPCT).
Since ABCD is a parallelogram, consecutive interior angles are supplementary.
So, \( \angle ABC + \angle BAD = 180^\circ \).
As \( \angle ABC = \angle BAD \), we have \( \angle BAD + \angle BAD = 180^\circ \).
\( \implies 2\angle BAD = 180^\circ \)
\( \angle BAD = 90^\circ \).
A rhombus with one angle equal to \( 90^\circ \) is a square.
Thus, ABCD is a square.
In simple words: If a four-sided shape has two main lines that are the same length, cut each other in half, and cross at a right angle, then that shape is a square. We prove this by showing it's first a parallelogram, then a rhombus, and finally a rectangle, which together means it's a square.

Exam Tip: To prove a quadrilateral is a square, you need to show it has the properties of both a rhombus (all sides equal) and a rectangle (all angles \( 90^\circ \)). Diagonals being equal and bisecting at right angles is the defining condition for a square.

Question 6. Diagonal AC of a parallelogram ABCD bisects \( \angle A \). Show that
(i) it bisects \( \angle C \) also,
(ii) ABCD is a rhombus
Answer:

Given: ABCD is a parallelogram and diagonal AC bisects \( \angle A \), meaning \( \angle DAC = \angle BAC \) (let's call them \( \angle 1 \) and \( \angle 2 \) respectively, so \( \angle 1 = \angle 2 \)).
To Prove: (i) AC bisects \( \angle C \) (i.e., \( \angle BCA = \angle DCA \)) and (ii) ABCD is a rhombus.
Proof:
**(i) AC bisects \( \angle C \):**
In parallelogram ABCD, \( AB || DC \).
So, \( \angle BAC = \angle DCA \) (Alternate interior angles). Let's call this \( \angle 2 = \angle 3 \). (Equation 1)
Also, \( AD || BC \).
So, \( \angle DAC = \angle BCA \) (Alternate interior angles). Let's call this \( \angle 1 = \angle 4 \). (Equation 2)
We are given that AC bisects \( \angle A \), so \( \angle 1 = \angle 2 \). (Equation 3)
From (1) and (3), since \( \angle 2 = \angle 3 \) and \( \angle 1 = \angle 2 \), we get \( \angle 1 = \angle 3 \).
From (2) and (3), since \( \angle 1 = \angle 4 \) and \( \angle 1 = \angle 2 \), we get \( \angle 2 = \angle 4 \).
Combining \( \angle 1 = \angle 3 \) and \( \angle 2 = \angle 4 \), with \( \angle 1 = \angle 2 \), we have \( \angle 3 = \angle 4 \).
This means \( \angle DCA = \angle BCA \).
Hence, diagonal AC also bisects \( \angle C \).

**(ii) ABCD is a rhombus:**
From our proof in (i), we found that \( \angle 1 = \angle 3 \) (i.e., \( \angle DAC = \angle BCA \)).
In \( \triangle ABC \), since \( \angle BAC = \angle BCA \) (as \( \angle 2 = \angle 4 \) and \( \angle 1 = \angle 2 \) implies \( \angle 1 = \angle 4 \), which means \( \angle BAC = \angle BCA \)), the sides opposite to these equal angles must be equal.
So, \( BC = AB \).
Since ABCD is a parallelogram and its adjacent sides \( AB \) and \( BC \) are equal, it must be a rhombus.
Therefore, ABCD is a rhombus.
In simple words: If a parallelogram's main line (diagonal) cuts one of its corner angles in half, then it must also cut the opposite corner angle in half. Also, this type of parallelogram will have all its sides equal, making it a rhombus.

Exam Tip: A key property of a rhombus is that its diagonals bisect the angles. If a parallelogram has a diagonal that bisects an angle, it's a strong indicator that the parallelogram is a rhombus.

Question 7. ABCD is a rhombus. Show that diagonal AC bisects \( \angle A \) as well as \( \angle C \) and diagonal BD bisects \( \angle B \) as well as \( \angle D \).
Answer:

Given: Rhombus ABCD, with AC and BD as diagonals.
To Prove: (i) Diagonal AC bisects \( \angle A \) and \( \angle C \).
(ii) Diagonal BD bisects \( \angle B \) and \( \angle D \).
Proof:
**(i) Diagonal AC bisects \( \angle A \) and \( \angle C \):**
In \( \triangle ADC \), since ABCD is a rhombus, \( AD = DC \) (sides of a rhombus are equal).
Angles opposite to equal sides in a triangle are equal. So, \( \angle DAC = \angle DCA \). (Equation 1)
Since ABCD is a rhombus, it is also a parallelogram, so \( AB || DC \).
Therefore, \( \angle BAC = \angle DCA \) (Alternate interior angles). (Equation 2)
From (1) and (2), we have \( \angle DAC = \angle BAC \). This shows that AC bisects \( \angle A \).
Similarly, from (1) and (2), we also have \( \angle DCA = \angle BAC \). And we know \( \angle BCA \) (alternate interior to \( \angle DAC \)) is equal to \( \angle DAC \). So \( \angle BCA = \angle DCA \). This proves AC bisects \( \angle C \).

**(ii) Diagonal BD bisects \( \angle B \) and \( \angle D \):**
In \( \triangle BCD \), \( BC = CD \) (sides of a rhombus).
So, \( \angle BDC = \angle DBC \) (Angles opposite to equal sides). (Equation 3)
Since \( AB || DC \), \( \angle ABD = \angle BDC \) (Alternate interior angles). (Equation 4)
From (3) and (4), we get \( \angle DBC = \angle ABD \). This proves BD bisects \( \angle B \).
Similarly, since \( AD || BC \), \( \angle ADB = \angle DBC \) (Alternate interior angles). Combining with \( \angle BDC = \angle DBC \), we get \( \angle ADB = \angle BDC \). This proves BD bisects \( \angle D \).
In simple words: When you have a rhombus, the main lines (diagonals) that connect opposite corners do a special job. Each diagonal cuts the two corner angles it passes through exactly in half. For instance, diagonal AC cuts angle A into two equal parts and also cuts angle C into two equal parts. The same happens with diagonal BD for angles B and D.

Exam Tip: This is a fundamental property of a rhombus. Always remember that its diagonals not only bisect each other at right angles but also bisect the vertex angles through which they pass.

Question 8. ABCD is a rectangle in which diagonal AC bisects \( \angle A \) as well \( \angle C \). Show that:
(i) ABCD is a square.
(ii) Diagonal BD bisects \( \angle B \) as well as \( \angle D \).
Answer:

Given: ABCD is a rectangle, and diagonal AC bisects \( \angle A \) (meaning \( \angle BAC = \angle DAC \)) and \( \angle C \) (meaning \( \angle BCA = \angle DCA \)).
To Prove: (i) ABCD is a square, and (ii) Diagonal BD bisects \( \angle B \) and \( \angle D \).
Proof:
**(i) ABCD is a square:**
Since ABCD is a rectangle, \( AB || DC \).
Therefore, \( \angle BAC = \angle DCA \) (Alternate interior angles). (Equation 1)
We are given that AC bisects \( \angle A \), so \( \angle BAC = \angle DAC \). (Equation 2)
From (1) and (2), we get \( \angle DAC = \angle DCA \).
In \( \triangle ADC \), since \( \angle DAC = \angle DCA \), the sides opposite to these angles must be equal.
So, \( AD = CD \).
Since ABCD is a rectangle and its adjacent sides \( AD \) and \( CD \) are equal, it means all sides are equal (\( AB = BC = CD = DA \)).
A rectangle with all sides equal is a square.
Hence, ABCD is a square.

**(ii) Diagonal BD bisects \( \angle B \) as well as \( \angle D \):**
Now that we have established ABCD is a square, we know it is also a rhombus.
From Question 7, we proved that in a rhombus, the diagonals bisect the angles.
Therefore, diagonal BD bisects \( \angle B \) and also bisects \( \angle D \).
*Alternatively, a direct proof:*
Since ABCD is a square, \( AB = AD \).
In \( \triangle ABD \), \( \angle ADB = \angle ABD \) (Angles opposite to equal sides). (Equation 3)
Also, since \( AB || DC \), \( \angle ABD = \angle BDC \) (Alternate interior angles). (Equation 4)
From (3) and (4), we get \( \angle ADB = \angle BDC \). This shows BD bisects \( \angle D \).
Similarly, since \( AD || BC \), \( \angle ADB = \angle DBC \) (Alternate interior angles). (Equation 5)
From (3) and (5), we get \( \angle ABD = \angle DBC \). This shows BD bisects \( \angle B \).
In simple words: If a rectangle's diagonal cuts one of its corner angles exactly in half, then that rectangle is actually a square. And because it's a square, its other diagonal will also cut its two angles in half.

Exam Tip: When proving a figure is a square, remember to demonstrate both the "rhombus properties" (equal sides) and "rectangle properties" (right angles). If a rectangle has a diagonal bisecting an angle, it implies equal adjacent sides, making it a square.

Question 9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that \( DP = BQ \). Show that
(i) \( \triangle APD \cong \triangle CQB \)
(ii) \( AP = CQ \)
(iii) \( \triangle AQB \cong \triangle CPD \)
(iv) \( AQ = CP \)
(v) APCQ is a parallelogram.
Answer:

Given: ABCD is a parallelogram, P and Q are points on diagonal BD such that \( DP = BQ \).
To Prove: (i) \( \triangle APD \cong \triangle CQB \), (ii) \( AP = CQ \), (iii) \( \triangle AQB \cong \triangle CPD \), (iv) \( AQ = CP \), (v) APCQ is a parallelogram.
Proof:
**(i) \( \triangle APD \cong \triangle CQB \):**
Consider \( \triangle APD \) and \( \triangle CQB \).
1. \( AD = CB \) (Opposite sides of a parallelogram are equal)
2. \( DP = BQ \) (Given)
3. \( \angle ADP = \angle CBQ \) (Alternate interior angles, since \( AD || BC \) and BD is a transversal)
Therefore, \( \triangle APD \cong \triangle CQB \) (by SAS congruence rule).

**(ii) \( AP = CQ \):**
Since \( \triangle APD \cong \triangle CQB \) (Proved above),
Their corresponding parts are equal. So, \( AP = CQ \) (by CPCT).

**(iii) \( \triangle AQB \cong \triangle CPD \):**
Consider \( \triangle AQB \) and \( \triangle CPD \).
1. \( AB = CD \) (Opposite sides of a parallelogram are equal)
2. \( BQ = DP \) (Given)
3. \( \angle ABQ = \angle CDP \) (Alternate interior angles, since \( AB || DC \) and BD is a transversal)
Therefore, \( \triangle AQB \cong \triangle CPD \) (by SAS congruence rule).

**(iv) \( AQ = CP \):**
Since \( \triangle AQB \cong \triangle CPD \) (Proved above),
Their corresponding parts are equal. So, \( AQ = CP \) (by CPCT).

**(v) APCQ is a parallelogram:**
From (ii), we have \( AP = CQ \).
From (iv), we have \( AQ = CP \).
Since both pairs of opposite sides of quadrilateral APCQ are equal, APCQ is a parallelogram.
In simple words: Inside a parallelogram, if you pick two points on a diagonal that are equally distant from the ends, then connecting these points to the other corners forms two pairs of identical triangles. This also means the lines AP, CQ, AQ, CP create another parallelogram inside the first one.

Exam Tip: For problems involving parallelograms and internal points, use the properties of parallelograms (opposite sides parallel and equal, alternate interior angles equal) and congruence rules (SAS, SSS, ASA) to prove relationships between triangles and other quadrilaterals.

Question 10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD respectively (See figure). Show that
(i) \( \triangle APB \cong \triangle CQD \)
(ii) \( AP = CQ \)
Answer:

Given: ABCD is a parallelogram. \( AP \perp BD \) and \( CQ \perp BD \).
To Prove: (i) \( \triangle APB \cong \triangle CQD \) and (ii) \( AP = CQ \).
Proof:
**(i) \( \triangle APB \cong \triangle CQD \):**
Consider \( \triangle APB \) and \( \triangle CQD \).
1. \( AB = CD \) (Opposite sides of a parallelogram are equal)
2. \( \angle APB = \angle CQD \) (Both are \( 90^\circ \) because AP and CQ are perpendiculars to BD)
3. \( \angle ABP = \angle CDQ \) (Alternate interior angles, since \( AB || DC \) and BD is a transversal)
Therefore, \( \triangle APB \cong \triangle CQD \) (by AAS congruence rule).

**(ii) \( AP = CQ \):**
Since \( \triangle APB \cong \triangle CQD \) (Proved above),
Their corresponding parts are equal. So, \( AP = CQ \) (by CPCT).
In simple words: In a parallelogram, if you draw straight lines from two opposite corners down to a diagonal so that they meet it at a right angle, then the two triangles formed by these lines and the parallelogram sides are identical. This also means the two perpendicular lines you drew are exactly the same length.

Exam Tip: When dealing with perpendiculars in parallelograms, look for right angles and parallel lines to identify alternate interior angles. AAS congruence is very useful in such proofs.

Question 11. In \( \triangle ABC \) and \( \triangle DEF \), \( AB = DE \), \( AB || DE \), \( BC = EF \) and \( BC || EF \), vertices A, B and C are joined to vertices D, E and F respectively (see figure). Show that
(i) Quadrilateral ABED is a parallelogram.
(ii) Quadrilateral BEFC is a parallelogram.
(iii) \( AD || CF \) and \( AD = CF \).
(iv) Quadrilateral ACFD is a parallelogram.
(v) \( AC = DF \)
(vi) \( \triangle ABC \cong \triangle DEF \)
Answer:

Given: In \( \triangle ABC \) and \( \triangle DEF \), \( AB = DE \), \( AB || DE \), \( BC = EF \), and \( BC || EF \). Vertices A, B, C are joined to D, E, F respectively.
Proof:
**(i) Quadrilateral ABED is a parallelogram:**
We are given \( AB = DE \) and \( AB || DE \).
If one pair of opposite sides of a quadrilateral is equal and parallel, then it is a parallelogram.
Therefore, ABED is a parallelogram.

**(ii) Quadrilateral BEFC is a parallelogram:**
We are given \( BC = EF \) and \( BC || EF \).
If one pair of opposite sides of a quadrilateral is equal and parallel, then it is a parallelogram.
Therefore, BEFC is a parallelogram.

**(iii) \( AD || CF \) and \( AD = CF \):**
From (i), ABED is a parallelogram.
This means \( AD = BE \) and \( AD || BE \) (Opposite sides of a parallelogram). (Equation 1)
From (ii), BEFC is a parallelogram.
This means \( BE = CF \) and \( BE || CF \) (Opposite sides of a parallelogram). (Equation 2)
From (1) and (2):
Since \( AD = BE \) and \( BE = CF \), it follows that \( AD = CF \).
Since \( AD || BE \) and \( BE || CF \), it follows that \( AD || CF \).

**(iv) Quadrilateral ACFD is a parallelogram:**
From (iii), we have proved that \( AD = CF \) and \( AD || CF \).
Since one pair of opposite sides of quadrilateral ACFD is equal and parallel, it is a parallelogram.

**(v) \( AC = DF \):**
From (iv), ACFD is a parallelogram.
Therefore, \( AC = DF \) (Opposite sides of a parallelogram are equal).

**(vi) \( \triangle ABC \cong \triangle DEF \):**
Consider \( \triangle ABC \) and \( \triangle DEF \).
1. \( AB = DE \) (Given)
2. \( BC = EF \) (Given)
3. \( AC = DF \) (Proved in (v))
Therefore, \( \triangle ABC \cong \triangle DEF \) (by SSS congruence rule).
In simple words: This problem asks us to show several things about two triangles and the quadrilaterals formed by connecting their matching corners. First, we show that ABED and BEFC are parallelograms because they each have one pair of opposite sides that are both equal in length and parallel. Then, because these are parallelograms, we deduce that AD is equal and parallel to CF. This makes ACFD also a parallelogram, which means AC and DF are equal. Finally, with all three sides of triangle ABC being equal to the corresponding three sides of triangle DEF, the two triangles are identical.

Exam Tip: This question tests your understanding of the conditions for a quadrilateral to be a parallelogram. The key is that if one pair of opposite sides is both equal and parallel, the figure is a parallelogram. Also, remember SSS congruence rule for triangles.

Question 12. ABCD is a trapezium in which \( AB || CD \) and \( AD = BC \) (see figure). Show that
(i) \( \angle A = \angle B \)
(ii) \( \angle C = \angle D \)
(iii) \( \triangle ABC \cong \triangle BAD \)
(iv) Diagonal AC = diagonal BD.
Answer:

Given: ABCD is a trapezium with \( AB || CD \) and \( AD = BC \).
Construction: Extend AB and draw a line through C parallel to DA, intersecting AB produced at E.

Proof:
Since \( AB || DC \) (given), and AE is an extension of AB, then \( AE || DC \).
By construction, \( AD || CE \).
Therefore, quadrilateral ADCE is a parallelogram (both pairs of opposite sides are parallel).

**(i) \( \angle A = \angle B \):**
In parallelogram ADCE, \( AD = CE \) (Opposite sides of a parallelogram).
We are given \( AD = BC \).
So, \( BC = CE \).
In \( \triangle BCE \), since \( BC = CE \), the angles opposite to these equal sides are equal.
Thus, \( \angle CBE = \angle CEB \).
Since ABCE is a straight line, \( \angle ABC + \angle CBE = 180^\circ \) (Linear pair axiom). (Equation 1)
Since ADCE is a parallelogram, \( \angle A + \angle CEB = 180^\circ \) (Consecutive interior angles, \( AD || CE \)). (Equation 2)
From (1), \( \angle CBE = 180^\circ - \angle ABC \).
From (2), \( \angle CEB = 180^\circ - \angle A \).
Since \( \angle CBE = \angle CEB \), we have \( 180^\circ - \angle ABC = 180^\circ - \angle A \).
This implies \( \angle ABC = \angle A \), or \( \angle B = \angle A \).

**(ii) \( \angle C = \angle D \):**
Since \( AB || DC \), the sum of consecutive interior angles is \( 180^\circ \).
So, \( \angle A + \angle D = 180^\circ \) (Equation 3)
And \( \angle B + \angle C = 180^\circ \) (Equation 4)
From (i), we know \( \angle A = \angle B \).
Substitute \( \angle B \) with \( \angle A \) in (4): \( \angle A + \angle C = 180^\circ \).
Now compare with (3): \( \angle A + \angle D = 180^\circ \) and \( \angle A + \angle C = 180^\circ \).
Therefore, \( \angle D = \angle C \).

**(iii) \( \triangle ABC \cong \triangle BAD \):**
Consider \( \triangle ABC \) and \( \triangle BAD \).
1. \( AB = BA \) (Common side)
2. \( BC = AD \) (Given)
3. \( \angle ABC = \angle BAD \) (Proved in (i) that \( \angle B = \angle A \))
Therefore, \( \triangle ABC \cong \triangle BAD \) (by SAS congruence rule).

**(iv) Diagonal AC = diagonal BD:**
Since \( \triangle ABC \cong \triangle BAD \) (Proved above),
Their corresponding parts are equal. So, \( AC = BD \) (by CPCT).
In simple words: If you have a trapezium where the two non-parallel sides are equal, then it's an isosceles trapezium. For this shape, we can show a few things: the base angles are equal (angle A equals angle B), the top angles are also equal (angle C equals angle D), and if you draw lines connecting opposite corners (diagonals), those lines will be the same length. Also, two specific triangles formed by these diagonals and sides will be identical.

Exam Tip: For an isosceles trapezium (where non-parallel sides are equal), the base angles are equal, opposite angles are supplementary, and diagonals are equal. These properties are often proven using a construction line to form a parallelogram or a triangle with equal sides.