ABCD is a quadrilateral in which AB||DC and AD=BC. Prove that angle A = angle B and angle C=angle D
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See diagram. You can see a general quadrilateral with 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 D
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 D
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