Question

In figure ABCD is a cyclic quadrilateral; O is the centre of the circle. If $\angle BOD={160}^{\circ }$, find the measure of $\angle BPD$ , as shown in the given figure.

Answer 1

Consider the arc BCD of the circle. This arc makes angle $\angle BOD={160}^{\circ }$ at the centre of the circle and $\angle BAD$ at a point A on the circumference.
$\therefore \angle BAD=\frac{1}{2}\angle BOD={80}^{\circ }$
Now, ABCD is a cyclic quadrilateral.
$\Rightarrow \angle BAD+\angle BCD={180}^{\circ }$
${80}^{\circ }+\angle BCD={180}^{\circ }$
$\Rightarrow \angle BCD={100}^{\circ }$
$\therefore \angle BPD=\angle BCD$ [$\angle BPDand\angle BCD$ are angles in the same segment]
$\Rightarrow \angle BPD={100}^{\circ }$