Question

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Answer 1

Answer:

Given: AB||CD and AD=BC

Construction:Draw perpendicular from A and B meeting DC at L and M respectively.

AB||CD and AL||BM(since sum of co-interior angles ALM and BML=180)

Hence, ABML is a parallelogram.

Furthermore, ABML is a rectangle.

Now in ΔADL and ΔBCM

AD=BC             (given)

AL=BM             (opposite sides of a rectangle)

∠ALD=∠BMC  (both 90°)

ΔADL≅ΔBCM (by RHS)

∠DAL=∠CBM

∠DAL+90°=∠CBM+90°

∠DAB=∠CBA

Hence,proved

Answer 2

Given:

In trapezium ABCD, AB // CD, and AD = BC

construction:

xy // AB // CD is drawn.

To Prove:

a) angle A = angle B

b) triangle ABC congruent to triangle BAD

Proof:

b) In triangle ABC and Triangle BAD

AB=AB. (common side)...............(i)

BY BASIC PROPORTIONALITY THEOREM.:

By/Cy = Ax/Dx............(ii)

BC=AD. (Given)............(iii)

From ( i), (ii)&(iii) triangle ABC congruent to triangle BAD .

a) angle A = angle B. (C.P.C.T)

Hence Proved