Question

In parallelogram ABCD, if the bisector of adjacent angles A and D intersect each other at point P, then the value of $\angle$ APD = ${60}^{\circ }$.

State with true or false.

• A. True
• B. False

Answer 1

The correct option is B False

According to the given statement, the figure will be shown below:


Since, AB is parallel to DC and AD is tranversal,
Therefore, $\angle$BAD + $\angle$ADC = ${180}^{\circ }$
Since, AP bisects $\angle$BAD
$\angle$PAD = $\frac{1}{2}$$\angle$ADC
Therefore, $\angle$PAD + $\angle$PDA = $\frac{1}{2}$ $\angle$BAD + $\frac{1}{2}$ $\angle$ADC
= $\frac{1}{2}$ ($\angle$BAD + $\angle$ADC)
= $\frac{1}{2}$ $\times$ ${180}^{\circ }$
= ${90}^{\circ }$
In $\Delta APD$
$\angle$PAD + $\angle$PDA + $\angle$APD = ${180}^{\circ }$ [sum of the angle of triangle]
${90}^{\circ }$ + $\angle$APD =${180}^{\circ }$
$\angle$APD = ${180}^{\circ }$ - ${90}^{\circ }$
= ${90}^{\circ }$