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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. $$\begin{array}{rl}⟺|SM|\cdot |SN|& =|SP|\cdot |SQ|\\ ⟺|SX{|}^2& =(|SX|-|PX|)(|SX|+|QX|)\\ ⟺d^2& =(d-b)(d+c)\\ ⟺d(c-b)& =bc\\ ⟺2(s-b)(s-c)& =bc\\ ⟺a^2-(b-c{)}^2& =2bc\\ ⟺a^2& =b^2+c^2\\ ⟺\text{The triangle }ABC\text{ has a right angle at vertex }A.& \end{array}$$