Question

ABCD is trapezium in which AB|| CD. If AD=BC, show that $\angle A=\angle Band\angle C=\angle D$.

Answer 1

Given in trapezium ABCD, AB||CD and AD=BC
Draw perpendiculars DP and CQ on AB
Consider triangles APD and BQC
$\angle P=\angle Q={90}^{\circ }$
DP = CQ [Distance between parallel sides is same]
AD=BC (Given)
Therefore, $\Delta APD\cong \Delta BQC$ [By RHS congruence criterion]
Hence $\angle A=\angle B$ and $\angle C=\angle D$ [Since, CPCT]