Preselection tests for the full-time training

First solution. Let $△ABC$ be any triangle. Considering triangle $A{I}_{c}C$, we have $$\begin{array}{rl}\measuredangle C{I}_{c}A& =180^{\circ }-\left(\measuredangle AC{I}_c+\measuredangle I_{c}AC\right)\\ & =180^{\circ }-\left(\frac{1}{2}\measuredangle ACB+\measuredangle BAC+\frac{1}{2}\left(180^{\circ }-\measuredangle BAC\right)\right)\\ & =90^{\circ }-\frac{1}{2}(\measuredangle ACB+\measuredangle BAC)\\ & =\measuredangle I_{b}BA.\end{array}$$



On the other hand, $\measuredangle I_{c}AC=\measuredangle BA{I}_b$. We deduce that triangles $AB{I}_b,A{I}_{c}C$ are similar and therefore $$\frac{A{I}_c}{AB}=\frac{AC}{A{I}_b}\iff A{I}_b\cdot A{I}_c=AB\cdot AC.$$ Thus, the area of triangle $ABC$ is equal to $\frac{1}{2}A{I}_b\cdot A{I}_c\sin \measuredangle BAC$. Hence, triangle $△ABC$ is right at $A$ if and only if its area is equal to $\frac{1}{2}A{I}_b\cdot A{I}_c$.

Second solution. Let $△ABC$ be any triangle. Denote $a,b,c$ the lengths of the opposite sides to the vertices $A,B,C$ respectively, $s$ the semiperimeter, $r$ the inradius, $r_b,r_c$ the exradius opposite to the vertices $B,C$ respectively. We have $$A{I}_b^2=r_b^2+(s-c{)}^2\text{ and }A{I}_c^2=r_c^2+(s-b{)}^2.$$



But the area of the triangle $△ABC$ is equal to $$K=rs=r_b(s-b)=r_c(s-c)=\sqrt{s(s-a)(s-b)(s-c)},$$ and that $r_b{r}_c=s(s-a)$ (which follows from the above formulas for the area). Hence $A{I}_b^2\cdot A{I}_c^2=\left(r_b^2+(s-c{)}^2\right)\left(r_c^2+(s-b{)}^2\right)=s^2(s-a{)}^2+2K^2+(s-b{)}^2(s-c{)}^2$. Therefore, $K=\frac{1}{2}A{I}_b\cdot A{I}_c$ is equivalent to $$2K^2=s^2(s-a{)}^2+(s-b{)}^2(s-c{)}^2.$$ But $2K^2=2s(s-a)(s-b)(s-c)$. Hence, this is equivalent to $$((s-b)(s-c)-s(s-a){)}^2=0$$ which simplifies to $$a^2=b^2+c^2.$$