62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour

Denote the inscribed circle by $k$, its center by $S$ (fig. 19), and its angles by $\alpha ,\beta ,\gamma$. From symmetry, $KL\perp CP$ and $\angle LPC=\angle KPC$. Then with simple calculations we get $\angle MKB=90^{\circ }-\frac{1}{2}\beta$ and $\angle LKC=90^{\circ }-\frac{1}{2}\gamma \Rightarrow \angle MKL=90^{\circ }-\frac{1}{2}\alpha$. Similarly for other angles.



Fig. 19

$$\text{As }\angle KPC+\frac{1}{2}\gamma =\angle BKP=90^{\circ }-\frac{1}{2}\beta ,$$ $$\angle LPC=\angle KPC=90^{\circ }-\frac{1}{2}(\beta +\gamma )=\frac{1}{2}\alpha .$$

$$\angle LPC=\angle LPS=\angle LAS=\frac{1}{2}\alpha ,$$

the quadrilateral $LSPA$ is inscribed. As $\angle ALS=90^{\circ }$, we get $AP\perp CP\Rightarrow AP\parallel KL$.