Question

In a cyclic quadrilateral $ABCD$ side $AB$ and $DC$ produced to meet at point $P$ and $AD$ and $BC$ produced to meet at point $Q$ (both $P$ and $Q$ are outside the circle). Find the value of $\angle PTO$ where $PT$ and $QT$ are the bisectors of $\angle APD$ and $\angle AQB$ respectively.

Answer 1

$\Rightarrow \angle PCQ=180°-\angle 1$

$\Rightarrow \angle PCQ=\angle PCQ-(\angle 5+\angle 6)$

$=180°-\angle 1-(\angle 5+\angle 6)⟶\left(1\right)$

$\Rightarrow 2\angle 5=\angle 4-\angle 1$

$\Rightarrow 2\angle 6=\angle 2-\angle 1$

$\Rightarrow \angle 5+\angle 6=\frac{(\angle 2+\angle 4-2\angle 1)}{2}$

$\angle 2+\angle 4=180°$

$\Rightarrow (\angle 5+\angle 6)=90°-\angle 1$

$\Rightarrow \angle PTO=180°-\angle 1-90°+\angle 1=90°.$

Hence, the answer is solved.