Question

Solve this congruency question​

Answer 1

ANSWER:-

Given:

In a quadrilateral, AB||DC & P is the midpoint of BC.

AP & DC meet at Q.

To prove:

⚫AB= CQ

⚫DQ= DC+ AB

Proof:

In ∆BPA & ∆CPQ

AB||DC &

AB||DQ

⚫AQ is a transversal

∠1= ∠1 [Vertically opposite angle]

∠2 = ∠2 [Vertically opposite angle]

In ∆ABP,

$\alpha = 180 - ( \angle 1 + \angle 2)$

In ∆CQP,

$\alpha = 180 - ( \angle1 + \angle 2)$

BP = PC [P is the midpoint of BC]

∠ABP = ∠QCP

∠BPA = ∠CPQ [vertically opposite angle]

∆BPA = ∆CPQ [ASA congruence rule]

⚫AB= CQ [c.p.c.t]

&

⚫DQ= DC+ CQ

=) DQ = DC + AB [CQ = AB]

Hence,

Proved.

Hope it helps ☺️