5.4. Find the angle $C$ of triangle $A B C$, if vertex ' $A$ is equidistant from the centers of the excircles *) tangent to sides $A B$ and $B C$.

*) Excircle - a circle that is tangent to one side of the triangle and the extensions of the other two sides.
5.4. Answer. $\angle C=90^{\circ}$.

Let $\mathrm{O}_{2}$ be the center of the excircle that touches side $A B$ and the extensions of sides $C B$ and

![](https://cdn.mathpix.com/cropped/2024_05_21_20eb9f6ef2c568605a63g-45.jpg?height=375&width=490&top_left_y=447&top_left_x=143)

Fig. 5.4. $C A, O_{1}$ be the center of the excircle that touches side $B C$ and the extensions of sides $A B$ and $A C$. Clearly, $O_{1}$ and $O_{2}$ are the points of intersection of the bisector of the external angle $B$ of triangle $A B C$ with the bisectors of the internal and external angles $A$ (see figure). Since $\angle O_{1} A O_{2}=90^{\circ}$ (bisectors of adjacent angles are perpendicular) and by condition $A O_{1}=A O_{2}$, then $\angle A O_{1} O_{2}=\angle A O_{2} O_{1}=45^{\circ}$. Now prove that $\angle C=180^{\circ}-2 \angle B O_{2} A$. For this, drop perpendiculars from $O_{2}$ to $A C, B C$ and $A B$.