Croatian Mathematical Competitions

Let $|BC| = a$, $|CA| = b$ and $|AB| = c$ be the lengths of the sides of the triangle $ABC$ and let $s = \frac{1}{2}(a+b+c)$ be its semiperimeter. Without loss of generality we may assume $b < c$.
Let $S$ be the intersection of lines $BC$ and $MN$, and let $X$ be the point where the incircle of the triangle $ABC$ touches the side $\overline{BC}$.

Menelaus' theorem applied to the line $MN$ and the triangle $ABC$ gives
$$\frac{|AM|}{|BM|} \cdot \frac{|BS|}{|CS|} \cdot \frac{|CN|}{|AN|} = 1.$$
Since $|AM| = |AN| = s-a$, $|BM| = |BX| = s-b$ and $|CN| = |CX| = s-c$, we have
$$\frac{|AM|}{|BM|} \cdot \frac{|BX|}{|CX|} \cdot \frac{|CN|}{|AN|} = 1,$$
and hence
$$\frac{|BX|}{|CX|} = \frac{|BS|}{|CS|} = \frac{|SX| + |BX|}{|SX| - |CX|}.$$
Denoting $|SX| = d$ gives
$$\frac{s-b}{s-c} = \frac{d+(s-b)}{d-(s-c)},$$
i.e. $d(c-b) = 2(s-b)(s-c)$.
We also have $|PX| = |CX| + |CP| = (s-c) + (s-a) = b$, analogously $|QX| = c$ and $|SM| \cdot |SN| = |SX|^2$ (by the power of the point $S$ with respect to the incircle of the triangle $ABC$).
Finally, the following sequence of equivalent statements finishes the proof:
The quadrilateral $MNPQ$ is cyclic.
$$\begin{align*} \iff |SM| \cdot |SN| &= |SP| \cdot |SQ| \\ \iff |SX|^2 &= (|SX| - |PX|)(|SX| + |QX|) \\ \iff d^2 &= (d-b)(d+c) \\ \iff d(c-b) &= bc \\ \iff 2(s-b)(s-c) &= bc \\ \iff a^2 - (b-c)^2 &= 2bc \\ \iff a^2 &= b^2 + c^2 \\ \iff \text{The triangle } ABC \text{ has a right angle at vertex } A. \end{align*}$$