SELECTION and TRAINING SESSION

(Solution by A. Gaponenko, D. Voynov.) First, note that incircles of the triangles $QAC$ and $QBC$ touch $CQ$ at the same point $X$ (well-known fact). Hence $CF=CX=CT$. Also $AP=AT$, $BF=BD$. Now we have $$\angle TFD=180^{\circ }-\angle TFC-\angle BFD=$$ $$=180^{\circ }-(90^{\circ }-\frac{1}{2}\angle C)-(90^{\circ }-\frac{1}{2}\angle B)=$$ $$=\frac{1}{2}(\angle C+\angle B)=90^{\circ }-\frac{1}{2}\angle A=\angle APT.$$ It follows that $$\angle TFD+\angle TPD=\angle TFD+(180^{\circ }-\angle APT)=180^{\circ }$$ which finishes the proof.