Question

In the given figure, A,B,C,D are points on a circle centred at O.The lines AC and BD are extended to meet at P and lines AD and BC intersect at Q. Then prove that $\angle APB+\angle AQB=\angle AOB$.

Answer 1

Join AB.Let $\angle AOB=x$

$\angle ACB=\angle ADB=\frac{x}{2}$ (Angles in the same segment)

Now, in quadrilateral CPDQ

$\angle CPD+\angle PDQ+\angle DQC+\angle QCP={360}^{\circ }$

$\angle CPD+180-\frac{x}{2}+\angle DQC+180-\frac{x}{2}={360}^{\circ }$

$\angle CPD+\angle DQC={360}^{\circ }-(360-\frac{x}{2}-\frac{x}{2})$

$\Rightarrow \angle CPD+\angle AQB=x$ (Since $\angle DQC=\angle AQB$)

$\angle APB+\angle AQB=\angle AOB$