In the following figure, $O$ is the incentre of the triangle $ABC.$ $P$ and $Q$ are points of contact of the circle with the sides. $BO$ extended meets $PQ$ at $T.$
Show that $\angle ATB$ is a right angle.
I have made a few observations:
$BO$ clearly bisects $\angle B$.
This means that if we show that $\Delta ATB\sim \Delta OQB,$ we are done. But it is not clear to me how to establish this similarity.
How do I proceed from here?
Thank you
It is easier to prove that quadrilateral $AOTP$ is cyclic.
Reason: $\angle PTO=\angle TQB+\angle TBQ=\angle TQB+\angle B/2$.
But $\angle TQB=\angle TQO+\angle OQB=\angle TQO+90^\circ$
On the other hand, $\angle TQO=\frac 1{2}(180^\circ-\angle POQ)$ and $\angle POQ=180^\circ-C$ since quadrilateral $CPOQ$ is cyclic, having opposite right angles. Therefore, $\angle TQO=C/2$
Conclusion:
$\angle PTO=90^\circ+\angle C/2+\angle B/2=180^\circ-\angle A/2$,
thus proving that quadrilateral $AOTP$ is cyclic.
As a consequence,
$\angle ATO=\angle APO=90^\circ$.