Question

$ABCD$ is a parallelogram. If $P$ be a point on $CD$ such that $AP=AD$ , then the measure of $\angle PAB+\angle BCD$ is

• A. ${180}^{\circ }$
• B. ${225}^{\circ }$
• C. ${240}^{\circ }$
• D. ${135}^{\circ }$

Answer 1

The correct option is A ${180}^{\circ }$

$\angle PAB=\angle APD$ [alternate interior angles]

And as in $△APD$

$AP=AD$

so, $\angle D=\angle APD=\angle PAB$

So, $\angle PAB+\angle BCD=\angle D+\angle BCD=\angle D+\angle C={180}^{\circ }$ since $\angle C$ and $\angle D+$ alternate opposite angle for $AD||BC$

$\therefore \angle PAB+\angle BCD={180}^{\circ }$