Question

Given a trapezium ABCD in which $AB||CD$ and $AD=BC$. If $\angle C={76}^{\circ }$, then $\angle D$ equals

• A. ${14}^{\circ }$
• B. ${104}^{\circ }$
• C. ${76}^{\circ }$
• D. None of these

Answer 1

The correct option is C ${76}^{\circ }$

In trapezium ABCD, $AB\parallel CD$, $AD=BC$
Draw a perpendicular from A on CD to meet CD at M and a perendicular from B on CD to meet at N.
Now, in $△ADM$ and $△BNC$
$\angle AMD=\angle BNC$ (Each ${90}^o$)
$AM=BN$ (distance between parallel lines)
$AD=BC$ (Given)
Thus, $△ADM\cong △BCN$ (SAS rule)
Hence, $\angle D=\angle C={76}^{\circ }$ (by CPCT)