ABCD is cyclic quadrilateral with ∠ADC = 135^o, sides BA and CD produced meet at point P, and sides AD and BC produced meet at point Q. If ∠P : ∠Q = 2 : 1, find ∠P and ∠Q.

It is given that,

P : Q = 2 : 1

Let ∠P = 2x and ∠Q = x

Also, in the cyclic quadrilateral ABCD, we have

∠ADC$={135}^{\circ }$ 

From the cyclic quadrilateral property,
$\angle B={180}^{\circ }-\angle \mathit{ADC}$
$\Rightarrow \angle B={180}^{\circ }-{135}^{\circ }$
$\Rightarrow \angle B={45}^{\circ }$

Since, in a linear pair, the sum of angles = 180^o.
$\angle \mathit{ADC}+\angle \mathit{ADP}={180}^{\circ }$.
 $\Rightarrow \angle \mathit{ADP}={180}^{\circ }-\angle \mathit{ADC}$
 $\Rightarrow \angle \mathit{ADP}={180}^{\circ }-{135}^{\circ }$
$\Rightarrow \angle \mathit{ADP}={45}^{\circ }$

Now, in ΔAQB,
$\angle A+\angle Q+\angle B={180}^{\circ }$[Angle sum property of triangle]
$\Rightarrow \angle A+x+{45}^{\circ }={180}^{\circ }$
$\Rightarrow \angle A={180}^{\circ }-\left(x+{45}^{\circ }\right)$
$\Rightarrow \angle A=\left({135}^{\circ }-x\right) \dots ..\left(1\right)$ 

Now, in ΔADP,

Exterior angle $\mathrm{A}=\angle \mathrm{ADP}+\angle \mathrm{P}$

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

Equate the value of ∠A from equations (1) and (2), we get

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

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

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

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

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

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

So, $\mathbf{\angle }\mathbf{P}\mathbf{=}\mathbf{60}\mathbf{°}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }\mathbf{\angle }\mathbf{Q}\mathbf{=}\mathbf{30}\mathbf{°}$.