In triangle $ABC$, let $AB = AC$. The angle bisectors of $\angle CAB$ and $\angle ABC$ intersect the sides $BC$ and $CA$ at points $D$ and $E$, respectively. Let $K$ be the incenter of triangle $ADC$. Suppose that $\angle BEK = 45^{\circ}$. Determine all possible values of $\angle CAB$.

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Kornis Kristóf's solution. Let $D^{\prime}$ be the reflection of point $D$ over the angle bisector from $C$. If $O$ is the center of the incircle of $\triangle ABC$, $\angle ODC = 90^{\circ} \Rightarrow \angle ODK = 45^{\circ}$, since $DK$ bisects $\angle ODC$. However, both $O$ and $K$ lie on the angle bisector from $C$, so $\angle OD^{\prime}K = 45^{\circ}$ and $D^{\prime}, E, O, K$ are on the same side, thus $\angle BEK = 45^{\circ}$, which means $O, E, D^{\prime}$, and $K$ lie on a circle.

Therefore, $\angle KED^{\prime} = \angle KOD^{\prime}$. However, $\angle KOD^{\prime} = \angle KOD$, and if $\angle BAC = \alpha$, then

\[
\begin{gathered}
\angle EBC = \frac{90^{\circ} - \frac{\alpha}{2}}{2}, \quad \angle ECB = 90^{\circ} - \frac{\alpha}{2} \Rightarrow \\
\Rightarrow \angle KED^{\prime} = \angle BED - 45^{\circ} = 180^{\circ} - \frac{90^{\circ} - \frac{\alpha}{2}}{2} - \left(90^{\circ} - \frac{\alpha}{2}\right) - 45^{\circ}, \\
\angle KOD = 90^{\circ} - \left(\frac{90^{\circ} - \frac{\alpha}{2}}{2}\right).
\end{gathered}
\]

Since

\[
\angle KED^{\prime} = \angle KOD, \quad 45^{\circ} - \frac{90^{\circ} - \frac{\alpha}{2} - \alpha}{2} = 90^{\circ} - \frac{90^{\circ} - \frac{\alpha}{2}}{2},
\]

$45^{\circ} = \frac{\alpha}{2} \Rightarrow \alpha = 90^{\circ}$.
![](https://cdn.mathpix.com/cropped/2024_05_02_beef25324be5503f42fdg-1.jpg?height=424&width=1092&top_left_y=1010&top_left_x=511)

There might be a problem with the above calculation if $D^{\prime} = E$, as in this case we cannot talk about $\angle KED^{\prime}$. If $D^{\prime} = E$, then $CD = CE$, and in this case, if the base is $a$ and the leg is $b$,

\[
CD = \frac{a}{2}, \quad CE = \frac{a \cdot b}{a + b}, \quad \frac{a}{2} = \frac{a \cdot b}{a + b} \quad \Rightarrow \quad a^2 + ab = 2ab \quad \Rightarrow \quad a = b
\]

the triangle is equilateral. If, however, the triangle is equilateral, $OEC D$ is a kite, a tangential quadrilateral, so the incircle of $\triangle ACD$ also touches $BE$, thus

\[
\angle BEK = \frac{\angle BEC}{2}, \quad \text{but} \quad \angle BEC = 90^{\circ}
\]

so $\angle BEK$ is also $45^{\circ}$ in this case.

In summary, we have two solutions: $\angle CAB = 90^{\circ}$, or $\angle CAB = 60^{\circ}$.