Estonian Mathematical Olympiad

Fig. 52 Solution: Let $\angle BAC=2\alpha$, $\angle CBA=2\beta$, $\angle ACB=2\gamma$; then $\alpha +\beta +\gamma =90^{\circ }$. By the two tangents theorem, we have $AE=AF$ (Fig. 52), and because $\angle FAK=\alpha =\angle KAE$, the triangles $AEK$ and $AFK$ are equal. Thus also $KE=KF$.

On the other hand, $\angle EIA=90^{\circ }-\alpha =\angle AIF$, implying that $\angle KIE=180^{\circ }-\angle EIA=180^{\circ }-\angle AIF=\angle FIK$. This shows that the angles inscribed on equal chords $KE$ and $KF$ of the circumcircles of the triangles $KIE$ and $KIF$ are equal. Consequently, these circles have equal radii and angles inscribed on equal chords of these circles are always equal.

From the triangle $AIC$, we obtain $\angle KIQ=\angle KIC=\alpha +\gamma =90^{\circ }-\beta$ and $\angle FIP=\angle FIB=90^{\circ }-\beta$. Thus angles inscribed on the chords $KQ$ and $PF$ of the circumcircles of the triangles $KIE$ and $KIF$ are equal, which implies $KQ=PF$. By the two tangents theorem, we also have $BF=BD$. As $\angle FBP=180^{\circ }-\beta =\angle PBD$, the triangles $FBP$ and $DBP$ are equal. Hence also $PF=PD$. Altogether, we obtain $KQ=PD$.

Analogously, we get $PK=DQ$. Hence the quadrilateral $PDQK$ is a parallelogram and its diagonals $PQ$ and $DK$ bisect each other.