(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.