# Selina Concise Solutions for Chapter 17 Special Types of Quadrilaterals Class 8 ICSE Mathematics

**1. In parallelogram ABCD, ∠A = 3 times ∠B. Find all the angles of the parallelogram. In the same parallelogram, if AB = 5x – 7 and CD = 3x + 1 ; find the length of CD.**

**Solution**

∠A = 3 ∠B = 3x

AD || BC

∠A + ∠B = 180°

3x + x = 180°

⇒ 4x = 180°

⇒ x = 45°

∠B = 45°

∠A = 3x = 3 x 45 = 135°

and ∠B = ∠D = 45°

opposite angles of ||gm are equal.

∠A = ∠C = 135°

opposite sides of //gm are equal.

AB = CD

5x – 7 = 3x + 1

⇒ 5x – 3x = 1+7

⇒ 2x = 8

⇒ x = 4°

CD = 3 x 4 + 1 = 13

Hence, 135°, 45°, 135° and 45°; 13

**2. In parallelogram PQRS, ∠Q = (4x – 5)° and ∠S = (3x + 10)°. Calculate : ∠Q and ∠R.**

**Solution**

In parallelogram PQRS,

∠Q = (4x – 5)° and ∠S = (3x + 10)°

∠Q = ∠S

4x – 5 = 3x + 10

⇒ 4x – 3x = 10 + 5

⇒ x = 15

∠Q = 4x – 5 = 4×15 – 5 = 55°

Also ∠Q + ∠R = 180°

⇒ 55° + ∠R = 180°

∠R = 180° - 55° = 125°

∠Q = 55° ; ∠R = 125°

**3. In rhombus ABCD ;**

**(i) if ∠A = 74° ; find ∠B and ∠C.**

**(ii) if AD = 7.5 cm ; find BC and CD.**

**Solution**

AD || BC

∠A + ∠B = 180°

⇒ 74° + ∠B = 180°

⇒ ∠B = 180° – 74° = 106°

∠A = ∠C = 74°

Sides of Rhombus are equal.

BC = CD = AD = 7.5 cm

(i) ∠B = 106° ; ∠C = 74°

(ii) BC = 7.5 cm and CD = 7.5 cm

**4. In square PQRS :**

**(i) if PQ = 3x – 7 and QR = x + 3 ; find PS**

**(ii) if PR = 5x and QR = 9x – 8. Find QS**

**Solution**

(i) sides of square are equal.

⇒ 3x – 7 = x + 3

⇒ 3x – x = 3 + 7

⇒ 2x = 10

⇒ x = 5

PS = PQ = 3x – 7 = 3×5 – 7 =8

(ii) PR = 5x and QS = 9x – 8

PR = QS

5x = 9x – 8

⇒ 5x – 9x = -8

⇒ -4x = -8

⇒ x = 2

QS = 9x – 8 = 9 ×2 – 8 = 10

**5. ABCD is a rectangle, if ∠BPC = 124°**

**Calculate: (i) ∠BAP (ii) ∠ADP**

**Solution**

∠PBC = ∠PCB = x (say)

But ∠BPC + ∠PBC + ∠PCB = 180°

124° + x + x = 180°

⇒ 2x = 180° – 124°

⇒ 2x = 56°

⇒ x = 28°

∠PBC = 28°

But ∠PBC = ∠ADP

**[Alternate ∠s]**

∠ADP = 28°

Again ∠APB = 180° – 124° = 56°

Also PA = PB

∠BAP = 1/2 (180° – ∠APB)

= 1/2 ×(180°- 56°)

= 1/2 ×124° = 62°

Hence (i) ∠BAP = 62° (ii) ∠ADP = 28°

ABCD is Rhombus

AB = BC

∠BAC = ∠ACB

But ∠BAC = 38°

∠ACB = 38°

In ∆ABC,

∠ABC + ∠BAC + ∠ACB = 180°

⇒ ∠ABC + 38° + 38° = 180°

⇒ ∠ABC = 180° – 76° = 104°

But ∠ABC = ∠ADC

∠ADC = 104°

⇒ ∠DAC = ∠DCA ( AD = CD)

⇒ ∠DAC = 1/2 [180° – 104°]

⇒ ∠DAC = 1/2 x 76° = 38°

Hence (i) ∠ACB = 38° (ii) ∠DAC = 38° (iii) ∠ADC = 104°

∠DAC = ∠BCA

But ∠BCA = 35°

∠DAC = 35°

But ∠DAC = ∠ACD

Hence (i) ∠BAP = 62° (ii) ∠ADP = 28°

**6. ABCD is a rhombus. If ∠BAC = 38°, find :****(i) ∠ACB****(ii) ∠DAC****(iii) ∠ADC.****Solution**ABCD is Rhombus

**(Given)**AB = BC

∠BAC = ∠ACB

**(∠s opp. to equal sides)**But ∠BAC = 38°

**(Given)**∠ACB = 38°

In ∆ABC,

∠ABC + ∠BAC + ∠ACB = 180°

⇒ ∠ABC + 38° + 38° = 180°

⇒ ∠ABC = 180° – 76° = 104°

But ∠ABC = ∠ADC

**(opp. ∠s of rhombus)**∠ADC = 104°

⇒ ∠DAC = ∠DCA ( AD = CD)

⇒ ∠DAC = 1/2 [180° – 104°]

⇒ ∠DAC = 1/2 x 76° = 38°

Hence (i) ∠ACB = 38° (ii) ∠DAC = 38° (iii) ∠ADC = 104°

**7. ABCD is a rhombus. If ∠BCA = 35°. find ∠ADC.**

**Solution****Given:**Rhombus ABCD in which ∠BCA = 35°**To find :**∠ADC

**Proof :**AD || BC∠DAC = ∠BCA

**(Alternate ∠s)**But ∠BCA = 35°

**(Given)**∠DAC = 35°

But ∠DAC = ∠ACD

**(AD = CD)**& ∠DAC +∠ACD + ∠ADC = 180°

⇒ 35° + 35° + ∠ADC = 180°

⇒ ∠ADC = 180° – 70° = 110°

Hence, ∠ADC = 110°

∠PMS = 54° ; ∠QSR = 25° and ∠SQR = 30°

⇒ ∠PSQ = ∠SQR

But ∠SQR = 30°

∠PSQ = 30°

In ∆SMP,

∠PMS + ∠PSM + ∠MPS = 180° or 54° + 30° + ∠RPS = 180°

⇒ ∠RPS = 180°- 84° = 96°

Now ∠PRS + ∠RSQ = ∠PMS

∠PRS + 25° = 54°

⇒ ∠PRS = 54° – 25° = 29°

⇒ ∠PSR = ∠PSQ + ∠RSQ = 30° + 25° = 55°

Hence (i) ∠RPS = 96° (ii) ∠PRS = 29° (iii) ∠PSR = 55°

MD = MB

Also ∠ADB = ∠DBN

& ∠DML = ∠BMN

∆DML = ∆BMN

LM = MN

M is mid-point of LN.

Hence proved.

(ii) ∠B + ∠D = 180°

⇒ ∠A + ∠D = 180°

But ∠A = ∠B

∠B + ∠D = 180°

Similarly ∠A + ∠C = 180°

Hence the result.

(i)

AC = BD

& AC ⊥ BD

i.e., Diagonals of quadrilateral are equal and they are ⊥r to each other.

∴ ABCD is square.

(ii)

AC ⊥BD

But AC & BD are not equal

∴ ABCD is a Rhombus.

(iii)

AC = BD but Ac & BD are not ⊥r to each other.

∴ ABCD is a rectangle.

∠1 = ∠2

∠3 = ∠4

and AB = CD

∆COD = ∆AOB

OA = OC and OB = OD

Hence the result.

⇒ 35° + 35° + ∠ADC = 180°

⇒ ∠ADC = 180° – 70° = 110°

Hence, ∠ADC = 110°

**8. PQRS is a parallelogram whose diagonals intersect at M.****If ∠PMS = 54°, ∠QSR = 25° and ∠SQR = 30° ; find :****(i) ∠RPS****(ii) ∠PRS****(iii) ∠PSR.**

**Solution****Given :**||gm PQRS in which diagonals PR & QS intersect at M.∠PMS = 54° ; ∠QSR = 25° and ∠SQR = 30°

**To find :**(i) ∠RPS (ii) ∠PRS (iii) ∠PSR**Proof :**QR || PS⇒ ∠PSQ = ∠SQR

**(Alternate ∠s)**But ∠SQR = 30°

**(Given)**∠PSQ = 30°

In ∆SMP,

∠PMS + ∠PSM + ∠MPS = 180° or 54° + 30° + ∠RPS = 180°

⇒ ∠RPS = 180°- 84° = 96°

Now ∠PRS + ∠RSQ = ∠PMS

∠PRS + 25° = 54°

⇒ ∠PRS = 54° – 25° = 29°

⇒ ∠PSR = ∠PSQ + ∠RSQ = 30° + 25° = 55°

Hence (i) ∠RPS = 96° (ii) ∠PRS = 29° (iii) ∠PSR = 55°

**9. Given: Parallelogram ABCD in which diagonals AC and BD intersect at M.****Prove: M is mid-point of LN.**

**Solution****Proof:**Diagonals of //gm bisect each other.MD = MB

Also ∠ADB = ∠DBN

**(Alternate ∠s)**& ∠DML = ∠BMN

**(Vert. opp. ∠s)**∆DML = ∆BMN

LM = MN

M is mid-point of LN.

Hence proved.

**10. In an Isosceles-trapezium, show that the opposite angles are supplementary.**

**Solution****Given:**ABCD is isosceles trapezium in which AD = BC**To Prove :**(i) ∠A + ∠C = 180°(ii) ∠B + ∠D = 180°

**Proof:**AB || CD⇒ ∠A + ∠D = 180°

But ∠A = ∠B

**[Trapezium is isosceles]**∠B + ∠D = 180°

Similarly ∠A + ∠C = 180°

Hence the result.

**11. ABCD is a parallelogram. What kind of quadrilateral is it if :****(i) AC = BD and AC is perpendicular to BD?****(ii) AC is perpendicular to BD but is not equal to it?****(iii) AC = BD but AC is not perpendicular to BD?**

**Solution**(i)

**(Given)**& AC ⊥ BD

**(Given)**i.e., Diagonals of quadrilateral are equal and they are ⊥r to each other.

∴ ABCD is square.

(ii)

**(Given)**But AC & BD are not equal

∴ ABCD is a Rhombus.

(iii)

∴ ABCD is a rectangle.

**12. Prove that the diagonals of a parallelogram bisect each other.**

**Solution****Given:**||gm ABCD in which diagonals AC and BD bisect each other.**To Prove:**OA = OC and OB = OD**Proof:**AB || CD**(Given)**∠1 = ∠2

**(alternate ∠s)**∠3 = ∠4

**(alternate ∠s)**and AB = CD

**(opposite sides of //gm)**∆COD = ∆AOB

**(A.S.A. rule)**OA = OC and OB = OD

Hence the result.

**13. If the diagonals of a parallelogram are of equal lengths, the parallelogram is a rectangle. Prove it.**

**Solution****Given:**//gm ABCD in which AC = BD**To Prove :**ABCD is rectangle.**Proof :**In ∆ABC and ∆ABD

AB = AB

AC = BD

BC = AD

∆ABC = ∆ABD

∠A = ∠B

But AD // BC

∠A + ∠B = 180°

∠A = ∠B = 90°

Similarly, ∠D = ∠C = 90°

Hence, ABCD is a rectangle.

AB = AB

**(Common)**AC = BD

**(Given)**BC = AD

**(opposite sides of ||gm)**∆ABC = ∆ABD

**(S.S.S. Rule)**∠A = ∠B

But AD // BC

**(opp. sides of ||gm are ||)**∠A + ∠B = 180°

∠A = ∠B = 90°

Similarly, ∠D = ∠C = 90°

Hence, ABCD is a rectangle.

**14. In parallelogram ABCD, E is the mid-point of AD and F is the mid-point of BC. Prove that BFDE is a parallelogram.**

**Solution****Given:**//gm ABCD in which E and F are mid-points of AD and BC respectively.**To Prove:**BFDE is a ||gm.**Proof :**E is mid-point of AD.

DE = 1/2 AD

Also F is mid-point of BC

BF = 1/2 BC

But AD = BC

BF = DE

Again AD || BC

⇒ DE || BF

Now DE || BF and DE = BF

Hence BFDE is a ||gm.

(ii) DE bisects ∠ADC

(iii) ∠DEC = 90°

**(Given)**DE = 1/2 AD

Also F is mid-point of BC

**(Given)**BF = 1/2 BC

But AD = BC

**(opp. sides of ||gm)**BF = DE

Again AD || BC

⇒ DE || BF

Now DE || BF and DE = BF

Hence BFDE is a ||gm.

**15. In parallelogram ABCD, E is the mid-point of side AB and CE bisects angle BCD. Prove that :****(i) AE = AD,****(ii) DE bisects and ∠ADC and****(iii) Angle DEC is a right angle.**

**Solution****Given:**||gm ABCD in which E is mid-point of AB and CE bisects ZBCD.**To Prove :**(i) AE = AD(ii) DE bisects ∠ADC

(iii) ∠DEC = 90°

**Const.**Join DE**Proof :**(i) AB || CD

and CE bisects it.

∠1 = ∠3

But ∠1 = ∠2

From (i) & (ii)

∠2 = ∠3

BC = BE

But BC = AD

and BE = AE

AD = AE

∠4 = ∠5

But ∠5 = ∠6

⇒ ∠4 = ∠6

DE bisects ∠ADC.

Now AD // BC

⇒ ∠D + ∠C = 180°

⇒ 2∠6 + 2∠1 = 180°

DE and CE are bisectors.

∠6 + ∠1 = 180°/2

⇒ ∠6 + ∠1 = 90°

But ∠DEC + ∠6 + ∠1 = 180°

⇒ ∠DEC + 90° = 180°

⇒ ∠DEC = 180° – 90°

⇒ ∠DEC = 90°

Hence the result.

(i) ∠PSB + ∠SPB = 90°

(ii) ∠PBS = 90°

(iii) ∠ABC = 90°

(iv) ∠ADC = 90°

(v) ∠A = 9°

(vi) ABCD is a rectangle

**(Given)**and CE bisects it.

∠1 = ∠3

**(alternate ∠s) …(i)**But ∠1 = ∠2

**(Given) ...(ii)**From (i) & (ii)

∠2 = ∠3

BC = BE

**(sides opp. to equal angles)**But BC = AD

**(opp. sides of ||gm)**and BE = AE

**(Given)**AD = AE

∠4 = ∠5

**(∠s opp. to equal sides)**But ∠5 = ∠6

**(alternate ∠s)**⇒ ∠4 = ∠6

DE bisects ∠ADC.

Now AD // BC

⇒ ∠D + ∠C = 180°

⇒ 2∠6 + 2∠1 = 180°

DE and CE are bisectors.

∠6 + ∠1 = 180°/2

⇒ ∠6 + ∠1 = 90°

But ∠DEC + ∠6 + ∠1 = 180°

⇒ ∠DEC + 90° = 180°

⇒ ∠DEC = 180° – 90°

⇒ ∠DEC = 90°

Hence the result.

**16. In the following diagram, the bisectors of interior angles of the parallelogram PQRS enclose a quadrilateral ABCD.****Show that:****(i) ∠PSB + ∠SPB = 90°****(ii) ∠PBS = 90°****(iii) ∠ABC = 90°****(iv) ∠ADC = 90°****(v) ∠A = 90°****(vi) ABCD is a rectangle****Thus, the bisectors of the angles of a parallelogram enclose a rectangle.**

**Solution****Given:**In parallelogram ABCD bisector of angles P and Q, meet at A, bisectors of ∠R and ∠S meet at C. Forming a quadrilateral ABCD as shown in the figure.**To prove:**(i) ∠PSB + ∠SPB = 90°

(ii) ∠PBS = 90°

(iii) ∠ABC = 90°

(iv) ∠ADC = 90°

(v) ∠A = 9°

(vi) ABCD is a rectangle

**Proof:**In parallelogram PQRS,

PS || QR (opposite sides)

∠P + ∠Q = 180°

and AP and AQ are the bisectors of consecutive angles ∠P and ∠Q of the parallelogram

∠APQ + ∠AQP = 1/2 ×180° = 90°

But in ∆APQ,

∠A + ∠APQ + ∠AQP = 180°

⇒ ∠A + 90° = 180°

⇒ ∠A = 180° – 90°

(v) ∠A = 90°

Similarly PQ || SR

∠PSB + SPB = 90°

(ii) and ∠PBS = 90°

But, ∠ABC = ∠PBS

(iii) ∠ABC = 90°

Similarly we can prove that

(iv) ∠ADC = 90° and ∠C = 90°

(vi) ABCD is a rectangle

Hence proved.

(i) AX = YC

(ii) AX is parallel to YC

(iii) AXCY is a parallelogram

PS || QR (opposite sides)

∠P + ∠Q = 180°

and AP and AQ are the bisectors of consecutive angles ∠P and ∠Q of the parallelogram

∠APQ + ∠AQP = 1/2 ×180° = 90°

But in ∆APQ,

∠A + ∠APQ + ∠AQP = 180°

**(Angles of a triangle)**⇒ ∠A + 90° = 180°

⇒ ∠A = 180° – 90°

(v) ∠A = 90°

Similarly PQ || SR

∠PSB + SPB = 90°

(ii) and ∠PBS = 90°

But, ∠ABC = ∠PBS

**(Vertically opposite angles)**(iii) ∠ABC = 90°

Similarly we can prove that

(iv) ∠ADC = 90° and ∠C = 90°

(vi) ABCD is a rectangle

**(Each angle of a quadrilateral is 90°)**Hence proved.

**17. In parallelogram ABCD, X and Y are midpoints of opposite sides AB and DC respectively. Prove that:****(i) AX = YC****(ii) AX is parallel to YC****(iii) AXCY is a parallelogram.**

**Solution****Given:**In parallelogram ABCD, X and Y are the mid-points of sides AB and DC respectively AY and CX are joined**To prove:**

(ii) AX is parallel to YC

(iii) AXCY is a parallelogram

**Proof:**AB || DC and X and Y are the mid-points of the sides AB and DC respectively

(i) AX = YC

(ii) and AX || YC

(iii) AXCY is a parallelogram

Hence proved.

AN, NC, CM and MA are joined

(i) ∆DMC = ∆BNA and so CM = AN

(ii) ∆AMD = ∆CNB and so AM = CN

(iii) ANCM is a parallelogram

(i) In ∆DMC and ∆BNA.

CD = AB

DM = BN

∠CDM = ∠ABN

∆DMC = ∆BNA

CM = AN

Similarly, in ∆AMD and ∆CNB

AD = BC

DM = BN

∠ADM = ∠CBN

∆AMD = ∆CNB

AM = CN

(iii) CM = AN and AM = CN

ANCM is a parallelogram

Hence proved.

In rhombus ABCD, diagonals AC and BD bisect each other at 90°

∠BCD = 80°

Diagonals bisect the opposite angles also ∠BCD = ∠BAD

∠BAD = 80° and ∠ABC = ∠ADC = 180° – 80° = 100°

Diagonals bisect opposite angles

∠OCB or ∠PCB = 80°/2 = 40°

In ∆PCM,

Ext. CPD = ∠OCB + ∠PMC

⇒ 110° = 40° + x

⇒ x = 110° – 40° = 70°

and ∠ADO = 1/2 ∠ADC = 1/2 x 100° = 50°

Hence, x = 70° and y = 50°

ABCD is a parallelogram and AC is its diagonal which bisects the opposite angle

Opposite sides of a parallelogram are equal

3x + 14 = 2x + 25

⇒ 3x – 2x = 25 – 14

⇒ x = 11

∴ x = 11 cm

∠DCA = ∠CAB

y + 9° = 24

⇒ y = 24° – 9° = 15°

∠DAB = 3y° + 5° + 24° = 3×15 + 5 + 24° = 50° + 24° = 74°

(i) AX = YC

**(1/2 of opposite sides of a parallelogram)**(ii) and AX || YC

(iii) AXCY is a parallelogram

**(A pair of opposite sides are equal and parallel)**Hence proved.

**18. The given figure shows parallelogram ABCD. Points M and N lie in diagonal BD such that DM = BN.****Prove that:****(i) ∆DMC = ∆BNA and so CM = AN****(ii) ∆AMD = ∆CNB and so AM CN****(iii) ANCM is a parallelogram.**

**Solution****Given:**In parallelogram ABCD, points M and N lie on the diagonal BD such that DM = BNAN, NC, CM and MA are joined

**To prove:**(i) ∆DMC = ∆BNA and so CM = AN

(ii) ∆AMD = ∆CNB and so AM = CN

(iii) ANCM is a parallelogram

**Proof :**(i) In ∆DMC and ∆BNA.

CD = AB

**(opposite sides of ||gm ABCD)**DM = BN

**(given)**∠CDM = ∠ABN

**(alternate angles)**∆DMC = ∆BNA

**(SAS axiom)**CM = AN

**(c.p.c.t.)**Similarly, in ∆AMD and ∆CNB

AD = BC

**(opposite sides of ||gm)**DM = BN

**(given)**∠ADM = ∠CBN

**(alternate angles)**∆AMD = ∆CNB

**(SAS axiom)**AM = CN

**(c.p.c.t.)**(iii) CM = AN and AM = CN

**(proved)**ANCM is a parallelogram

**(opposite sides are equal)**Hence proved.

**19. The given figure shows a rhombus ABCD in which angle BCD = 80°. Find angles x and y.****Solution**In rhombus ABCD, diagonals AC and BD bisect each other at 90°

∠BCD = 80°

Diagonals bisect the opposite angles also ∠BCD = ∠BAD

**(Opposite angles of rhombus)**∠BAD = 80° and ∠ABC = ∠ADC = 180° – 80° = 100°

Diagonals bisect opposite angles

∠OCB or ∠PCB = 80°/2 = 40°

In ∆PCM,

Ext. CPD = ∠OCB + ∠PMC

⇒ 110° = 40° + x

⇒ x = 110° – 40° = 70°

and ∠ADO = 1/2 ∠ADC = 1/2 x 100° = 50°

Hence, x = 70° and y = 50°

**20. Use the information given in the alongside diagram to find the value of x, y and z.****Solution**ABCD is a parallelogram and AC is its diagonal which bisects the opposite angle

Opposite sides of a parallelogram are equal

3x + 14 = 2x + 25

⇒ 3x – 2x = 25 – 14

⇒ x = 11

∴ x = 11 cm

∠DCA = ∠CAB

**(Alternate angles)**y + 9° = 24

⇒ y = 24° – 9° = 15°

∠DAB = 3y° + 5° + 24° = 3×15 + 5 + 24° = 50° + 24° = 74°

∠ABC = 180° - ∠DAB = 180° – 74° = 106°

z = 106°

Hence, x = 11 cm, y = 15°, z = 106°

ABCD is a rectangle,

x : y = 3 : 1

In ∆BCE, ∠B = 90°

x + y = 90°

But x : y = 3 : 7

Sum of ratios = 3 + 7 = 10

∴ x = (90˚× 3)/10 = 27˚

And y = (90˚ × 7)/10 = 63˚

Hence, x = 27˚, y = 63˚

ABCE is a quadrilateral in which AC is its diagonal and AB || EC, AB = AC

BA is produced to D

AE bisects ∠DAC

(i) ∠EAC = ∠ACB

(ii) ABCE is a parallelogram

(i) In ∆ABC and ∆ZAEC

AC=AC

AB = CE

∠BAC = ∠ACE

∆ABC = ∆AEC

(ii) ∠BCA = ∠CAE

But these are alternate angles

AE || BC

But AB || EC

∴ ABCD is a parallelogram

Hence, x = 11 cm, y = 15°, z = 106°

**21. The following figure is a rectangle in which x: y = 3 : 7; find the values of x and y.****Solution**ABCD is a rectangle,

x : y = 3 : 1

In ∆BCE, ∠B = 90°

x + y = 90°

But x : y = 3 : 7

Sum of ratios = 3 + 7 = 10

∴ x = (90˚× 3)/10 = 27˚

And y = (90˚ × 7)/10 = 63˚

Hence, x = 27˚, y = 63˚

**22. In the given figure, AB // EC, AB = AC and AE bisects ∠DAC. Prove that:****(i) ∠EAC = ∠ACB****(ii) ABCE is a parallelogram.**

**Solution**ABCE is a quadrilateral in which AC is its diagonal and AB || EC, AB = AC

BA is produced to D

AE bisects ∠DAC

**To prove:**(i) ∠EAC = ∠ACB

(ii) ABCE is a parallelogram

**Proof:**(i) In ∆ABC and ∆ZAEC

AC=AC

**(common)**AB = CE

**(given)**∠BAC = ∠ACE

**(Alternate angle)**∆ABC = ∆AEC

**(SAS Axiom)**(ii) ∠BCA = ∠CAE

**(c.p.c.t.)**But these are alternate angles

AE || BC

But AB || EC

**(given)**∴ ABCD is a parallelogram