Question

$ABCD$ is a cyclic quadrilateral with $\angle ADC={135}^o$. Sides $BA$ and $CD$ produced meet at point $P$, sides $AD$ and $BC$ produced meet at point $Q$. If $\angle P:\angle Q=2:1$, find the angles $P$ and $Q$.

Answer 1

Let $\angle P=2x$ then $\angle Q=x$

Now,in cyclic quadrilateral $ABCD$,

$\angle B={180}^{\circ }-\angle ADC={180}^{\circ }-{135}^{\circ }={45}^{\circ }$

And $\angle ADC+\angle ADP={180}^{\circ }$(linear pair)

$\Rightarrow \angle ADP={45}^{\circ }$

Now,in $△AQB,$

$\angle A={180}^{\circ }-({45}^{\circ }+x)={135}^{\circ }-x$ ......$\left(1\right)$

In $△ADP,$

Exterior $\angle A=\angle ADP+\angle P={45}^{\circ }+2x$ ........$\left(2\right)$

From $\left(1\right)$ and $\left(2\right),$

${45}^{\circ }+2x={135}^{\circ }-x$

$\Rightarrow 3x={90}^{\circ }$

$\Rightarrow x={30}^{\circ }$

$\Rightarrow \angle P=2x=2\times {30}^{\circ }={60}^{\circ }$ and $\angle Q=x={30}^{\circ }$