Question

ABCD is cyclic quadrilateral with $\angle ADC=135{}^\circ $, 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 angles P and Q.

Answer 1

Hint: In a cyclic quadrilateral the sum of opposite angles of the quadrilateral is always equal to $180{}^\circ $. Also, in any triangle the sum of its three angles is always equal to $180{}^\circ $.

Complete step-by-step answer:

An exterior angle of a triangle is equal to the sum of the opposite interior angles.

Given:

$P:Q=2:1$

Let the ratio constant be x

$\Rightarrow \dfrac{P}{Q}=\dfrac{2x}{1x}$

$\Rightarrow $ Hence, $\angle P=2x$, then $\angle Q=x$

Now, in cyclic quadrilateral ABCD,

Given, $\angle ADC=135{}^\circ $

From the property of cyclic quadrilateral (given in hint)

$\begin{align}

  & \angle B=180{}^\circ -\angle ADC \\

 & \Rightarrow \angle B=180{}^\circ -135{}^\circ \\

 & \Rightarrow \angle B=45{}^\circ \\

\end{align}$

We know the sum of angles of a linear pair is equal to $180{}^\circ $.

Hence, $\angle ADC+\angle ADP=180{}^\circ $

$\begin{align}

  & \Rightarrow \angle ADP=180{}^\circ -\angle ADC \\

 & \Rightarrow \angle ADP=180{}^\circ -135{}^\circ \\

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

\end{align}$

Now, in $\Delta AQB$,

From the property of sum of angles of a triangle (given in hint)

$\begin{align}

  & \Rightarrow \angle A+\angle Q+\angle B=180{}^\circ \\

 & \Rightarrow \angle A+x+45{}^\circ =180{}^\circ \\

 & \Rightarrow \angle A=180{}^\circ -\left( x+45{}^\circ \right) \\

 & \Rightarrow \angle A=\left( 135{}^\circ -x \right)...........\left( i \right) \\

\end{align}$

 In $\Delta ADP$,

From exterior angle property of triangle (given in hint)

Exterior angle $A=\angle ADP+\angle P$

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

Equating (i) and (ii), we get,

$\Rightarrow 45{}^\circ +2x=135{}^\circ -x$ (Taking terms of x to one side)

$\Rightarrow 3x=135{}^\circ -45{}^\circ $

$\Rightarrow 3x=90{}^\circ $ (Dividing both sides of equation by 3)

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

Hence, $\angle Q=30{}^\circ $

$\angle P=2x=60{}^\circ $

Note: In this question, it is very important to draw a rough diagram before solving. In this question the relation between $\angle P$ and $\angle Q$ is given. We need to be careful while choosing the triangles. Therefore, we will choose $\Delta ABQ$ and $\Delta APD$ to apply Angle Sum Property and Exterior Angle Property in these triangles respectively to establish an equation to get the ratio constant.