China Hong Kong Mathematical Olympiad

a. Since $\angle FKI = \angle FKB = \angle EFA - \angle KBF = 90^\circ - \frac{A}{2} - \frac{B}{2} = \frac{C}{2} = \angle ECI$, the points $C$, $E$, $K$, $I$ are concyclic. Note that $C$, $E$, $I$, $D$ are also concyclic. This shows $C$, $E$, $K$, $I$, $D$ lie on the same circle. It follows that
$$\angle BKC = \angle IKC = \angle IEC = 90^\circ.$$

b. Let the midpoint of $CL$ be $J$. Applying Menelaus' theorem using the line $MPJ$ and $\triangle CKL$, we get $\frac{KP}{PC} \cdot \frac{CJ}{JL} \cdot \frac{LM}{MK} = 1$. Using $CJ = JL$, this simplifies

to
$$\frac{KP}{PC} = \frac{KM}{ML}. \qquad (1)$$
Now, note that $B$, $D$, $I$, $F$, $L$ are concyclic and $\angle BLC = 90^\circ$ as similar to the proof of part (a). Since $\angle BLC = \angle BKC = 90^\circ$, we know that $B$, $C$, $K$, $L$ are concyclic. Hence,
$$\angle QDL = \angle IDL = \angle IBL = \angle KBL = \angle KCL = \angle KCI = \angle KDI = \angle KDQ.$$
Together with $DQ \perp DM$, the lines $DQ$ and $DM$ are the internal and external angle bisectors of $\angle KDL$ respectively. So we get
$$\frac{KQ}{QL} = \frac{KD}{DL} = \frac{KM}{ML}. \qquad (2)$$
Combining (1) and (2), we obtain $\frac{KP}{PC} = \frac{KQ}{QL}$, which yields $PQ // CL$.