Question

if ABCD is a quadrilateral in which AB is parallel to CD and AD=BC , prove that angle A = angle B

Answer 1

AB || DC and AD = BC.

Extend the sides AD and BC till E and F as shown.  

As AB and CD are two parallel lines and AD intersects them both, the angles D and EAB are same.  But angle EAB = 180 - A.  so,

             angle D = 180 - A

Similarly, the line BC intersects parallel lines AB and DC, so

           angle C = angle FBA = 180 - B 

Now, draw perpendiculars from D and C onto AB meeting AB at G and H respectively.  

     Since AB || DC, the sides DG || CH.  

               Also,  DG = CH = distance between the parallel lines.

Looking at the triangles DGA and CHB, we find that

 

            DG = CH,    AD = BC (given),  angle G = angle H  = 90°.

  

  Δ DGA and Δ CHB are congruent.  Hence angle A = angle B.

 

     So, angle C = 180 - angle A = 180 - angle B = angle

         

hope it helps

Answer 2

Answer:

Step-by-step explanation:

Given: ABCD is a quadrilateral AB||CD and AD = BC.

To prove: ∠A = ∠B

Construction: Draw AP⊥CD and BQ⊥CD.

Proof: In ∆ADP and ∆BCQ,

AD = BC   (given)

AP = BQ   (distance between the parallel sides is equal)

∠APD = ∠BQC   (each is 90°)

So, ∆ADP ≅ ∆BCQ   (RHS congruence criterion)

⇒ ∠ADP = ∠BCQ  (C.P.C.T)

∠A + ∠ADP = 180°   (Sum of adjacent interior angle is supplementary)

∠B + ∠BCQ = 180°   (Sum of adjacent interior angle is supplementary)

⇒∠A + ∠ADP = ∠B + ∠BCQ

⇒∠A = ∠B (Since ∠ADP = ∠BCQ