Question

Consider the cyclic quadrilateral ABCD. If O is the centre of the circle and $\angle$BOD = ${160}^{\circ }$, then find the measure of $\angle$BPD.

(2 Marks)

Answer 1

We know that the angle subtended by an arc at the centre of a circle is double the angle subtended by it on any remaining part of the circle.

$⟹\angle BAD=\frac{\angle BOD}{2}={80}^{\circ }$ (1 mark)

Consider the cyclic quadrilateral ABPD. Since the opposite angles are supplementary,

$\angle BAD+\angle BPD={180}^{\circ }$

i.e., ${80}^{\circ }+\angle BPD={180}^{\circ }$

$⟹\angle BPD={100}^{\circ }$ (1 mark)