(Solution by V. Vityaz, A. Gaponenko.) Let without loss of generality $CX \ge CY$. Then $AP \le AN, BM \le BQ$ (see the Fig.) By the Menelaus theorem, we have: for the triangle $AQM$ and the line $XC$

$$\frac{AY}{YQ} \cdot \frac{QC}{CM} \cdot \frac{MX}{XA} = 1, \quad (1)$$

for the triangle $AQC$ and the line $NB$
$$\frac{AY}{YQ} \cdot \frac{QB}{BC} \cdot \frac{CN}{NA} = 1, \quad (2)$$

for the triangle $AMC$ and the line $PB$
$$\frac{AX}{XM} \cdot \frac{MB}{BC} \cdot \frac{CP}{PA} = 1. \qquad (3)$$
Let $AC = b$, $AB = c$, $BC = a$, $p$ be the semiperimeter of the triangle $ABC$. Then (2) and (3) give $AY/YQ = a/(p-b)$ and $MX/XA = 0.5(p-c)/(p-a)$, respectively. Substituting these equalities in (1) we get
$$\frac{a}{p-b} \cdot \frac{2(p-c)}{a} \cdot \frac{p-c}{2(p-a)} = 1 \iff (p-c)^2 = (p-a)(p-b) \iff$$
$$a^2 + b^2 = c(a+b) \iff a(p-c) = b(p-b) \iff \frac{YA}{YQ} = \frac{CA}{CQ},$$
the last proportion is equivalent to the fact that $CY$ bisects the angle $ACQ$.