Question

If in a quadrilateral, each pair of opposite angles are equal then it is a parallelogram.​

Answer 1

${\blue{\sf\underline{Given}}}$

A quad ABCD in which

$\angle \: A = \angle \: C \: and \: \angle \: B = \angle \: C$

${\blue{\sf\underline{To\:Prove}}}$

ABCD is a ||gm.

${\blue{\sf\underline{Proof\:\:\:We\:Have}}}$

$\angle \: A = \angle \: C \: and \: \angle \: B = \angle \: D \: \: (given)$

$⇒ \angle \: A + \angle \: B = \angle \: C \: + \angle \: D$

$⇒ \angle \: A + \angle \: B \: = \: \angle \: C \: + \angle \: D \: =180\degree \: \\\therefore \: \angle \: A \: + \angle \:B + \angle \: C + \angle \: D =360 \degree \brack \:$

Now, the the line segments BC and AD are cut by the transversal AB such that $\angle \: A + \angle \: B =180 \degree$

$\therefore \: AD || BC \\ \: \therefore \: \angle \: A \: and \: \angle \: B \: are \: co - interior \: \angle \: s\brack$

$again \: \angle \: A \: = \angle \: C \: and \: \angle \: D = \angle \: B \: ⇒ \angle \: A + \angle \: D = \angle \: C + \angle \: B$

$⇒ \angle \: A + \angle \: D = \angle \: C \: + \: \angle \: B \: = \: 180\degree \\ \therefore \: \angle \: A + \angle \: B + \angle \: C + \angle \: D \: = 360\degree\brack$

now the line segments DC and ab are cut by the transversal DA that $\angle \: A + \angle \: D \: = 180\degree$

$\therefore \: AB || DC \\\therefore \: \angle \: A \: and \: \angle \: D \: are \: co - interior \: \angle \: s \brack$

Thus,AB||DC and AD||BC.

Hence,ABCD is a ||gm.

Answer 2

$\implies$Qυєѕтıσи :

If in a quadrilateral, each pair of opposite angles are equal then it is a parallelogram.

$\implies$Aиѕωєя :

Given: A quadrilateral ABCD in which opposite angles are equal i.e., ∠A = ∠C ad ∠B = ∠D

To prove: ABCD is a parallelogram i.e, AB ║ DC and AD ║ BC.

Proof: Since, the sum of the angles of quadrilateral is 3600

⇒ ∠A + ∠B + ∠C + ∠D = 3600

⇒ ∠A + ∠D + ∠A + ∠D = 360

[∠A = ∠C and ∠B = ∠D]

⇒ 2∠A = 2∠D = 3600

⇒ ∠A + ∠D = 1800 [Co-interior angle]

⇒ AB ║ DC

∠A + ∠B + ∠C + ∠D = 3600

⇒ ∠A + ∠B + ∠A + ∠B = 3600 [∠A = ∠C and ∠B = ∠D] ⇒ 2∠A + 2∠B = 3600

⇒ ∠A + ∠B = 1800

[∵This is sum of interior angles on the same side of transversal AB]

∴ AD ║ BC So, AB ║ DC and AD ║ BC

$\implies$ABCD is a parallelogram.