First solution. Let $\triangle ABC$ be any triangle. Considering triangle $A I_{c} C$, we have
$$\begin{aligned} \measuredangle C I_{c} A & =180^{\circ}-\left(\measuredangle A C I_{c}+\measuredangle I_{c} A C\right) \\ & =180^{\circ}-\left(\frac{1}{2} \measuredangle A C B+\measuredangle B A C+\frac{1}{2}\left(180^{\circ}-\measuredangle B A C\right)\right) \\ & =90^{\circ}-\frac{1}{2}(\measuredangle A C B+\measuredangle B A C) \\ & =\measuredangle I_{b} B A . \end{aligned}$$

On the other hand, $\measuredangle I_{c} A C=\measuredangle B A I_{b}$. We deduce that triangles $A B I_{b}, A I_{c} C$ are similar and therefore
$$\frac{A I_{c}}{A B}=\frac{A C}{A I_{b}} \quad \Leftrightarrow \quad A I_{b} \cdot A I_{c}=A B \cdot A C .$$
Thus, the area of triangle $A B C$ is equal to $\frac{1}{2} A I_{b} \cdot A I_{c} \sin \measuredangle B A C$. Hence, triangle $\triangle ABC$ 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 $\triangle ABC$ 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
$$A I_{b}^{2}=r_{b}^{2}+(s-c)^{2} \quad \text{ and } \quad A I_{c}^{2}=r_{c}^{2}+(s-b)^{2} .$$

But the area of the triangle $\triangle ABC$ 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
$A I_{b}^{2} \cdot A I_{c}^{2}=\left(r_{b}^{2}+(s-c)^{2}\right)\left(r_{c}^{2}+(s-b)^{2}\right)=s^{2}(s-a)^{2}+2 K^{2}+(s-b)^{2}(s-c)^{2}$.
Therefore, $K=\frac{1}{2} A I_{b} \cdot A I_{c}$ is equivalent to
$$2 K^{2}=s^{2}(s-a)^{2}+(s-b)^{2}(s-c)^{2} .$$
But $2 K^{2}=2 s(s-a)(s-b)(s-c)$. Hence, this is equivalent to
$$((s-b)(s-c)-s(s-a))^{2}=0$$
which simplifies to
$$a^{2}=b^{2}+c^{2} .$$