Tokarev S.i.

In triangle $ABC$, the bisectors $AD$ and $BE$ are drawn. It is known that $DE$ is the bisector of angle $ADC$. Find the measure of angle $A$.
Prove that $E$ is the center of the excircle of triangle $ADB$.

## Solution

Point $E$ is equidistant from the lines $AD$, $BC$, and $AB$, since it lies on the angle bisectors $DE$ and $BE$ of angles $ADC$ and $ABC$. Therefore, $E$ is the center of the excircle of triangle $ADB$. Hence, point $E$ lies on the bisector of the external angle at vertex $A$ of triangle $ABD$. Since $AD$ is the bisector of angle $BAC$, the rays $AE$ and $AD$ divide the straight angle at vertex $A$ into three equal angles. Therefore, each of these angles is $60^{\circ}$, and $\angle A = 120^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_a69fc833ac69b1da7555g-50.jpg?height=363&width=829&top_left_y=1&top_left_x=640)

## Answer

$120^{\circ}$.