Question

Using ruler and compasses only, construct a parallelogram ABCD using the following data:
AB = $6$ cm, AD = $3$ cm and $\angle DAB={45}^{\circ }$. If the bisector of $\angle DAB$ meets DC at P, prove that $\angle APB$ is a right angle.

Answer 1

In the case of parallelogram, the opposite sides are equal. The sum of co-interior angles is $180°$ and the opposite angles are equal.

$\therefore \angle DAP$ bisect angle A,

$\therefore \angle ABP$ bisect angles B.

From $△APB$

$\Rightarrow \angle PAB+\angle PBA+\angle APB=180°$ [ Sum of triangle ppty ]

$\Rightarrow 22.5°+67.5°+\angle APB=180°$

$\Rightarrow \angle APB=90°$

Hence, the answer is $90°.$