BEL   Let $A B C$ be a triangle with $A B=A C$. The angle bisectors of $A$ and $B$ meet the sides $B C$ and $A C$ in $D$ and $E$, respectively. Let $K$ be the incenter of triangle $A D C$. Suppose that $\angle B E K=45^{\circ}$. Find all possible values of $\angle B A C$.
Answer: $\angle B A C=60^{\circ}$ or $\angle B A C=90^{\circ}$ are possible values and the only possible values. Let $I$ be the incenter of triangle $A B C$, then $K$ lies on the line $C I$. Let $F$ be the point, where the incircle of triangle $A B C$ touches the side $A C$; then the segments $I F$ and $I D$ have the same length and are perpendicular to $A C$ and $B C$, respectively. ![](https://cdn.mathpix.com/cropped/2024_11_18_b191c9407e8cb3d672cbg-49.jpg?height=549&width=615&top_left_y=1025&top_left_x=252)  Figure 1 ![](https://cdn.mathpix.com/cropped/2024_11_18_b191c9407e8cb3d672cbg-49.jpg?height=510&width=901&top_left_y=1067&top_left_x=906)  Figure 2  Let $P, Q$ and $R$ be the points where the incircle of triangle $A D C$ touches the sides $A D, D C$ and $C A$, respectively. Since $K$ and $I$ lie on the angle bisector of $\angle A C D$, the segments $I D$ and $I F$ are symmetric with respect to the line $I C$. Hence there is a point $S$ on $I F$ where the incircle of triangle $A D C$ touches the segment $I F$. Then segments $K P, K Q, K R$ and $K S$ all have the same length and are perpendicular to $A D, D C, C A$ and $I F$, respectively. So - regardless of the value of $\angle B E K$ - the quadrilateral $K R F S$ is a square and $\angle S F K=\angle K F C=45^{\circ}$. Consider the case $\angle B A C=60^{\circ}$ (see Figure 1). Then triangle $A B C$ is equilateral. Furthermore we have $F=E$, hence $\angle B E K=\angle I F K=\angle S E K=45^{\circ}$. So $60^{\circ}$ is a possible value for $\angle B A C$. Now consider the case $\angle B A C=90^{\circ}$ (see Figure 2). Then $\angle C B A=\angle A C B=45^{\circ}$. Furthermore, $\angle K I E=\frac{1}{2} \angle C B A+\frac{1}{2} \angle A C B=45^{\circ}, \angle A E B=180^{\circ}-90^{\circ}-22.5^{\circ}=67.5^{\circ}$ and $\angle E I A=\angle B I D=180^{\circ}-90^{\circ}-22.5^{\circ}=67.5^{\circ}$. Hence triangle $I E A$ is isosceles and a reflection of the bisector of $\angle I A E$ takes $I$ to $E$ and $K$ to itself. So triangle $I K E$ is symmetric with respect to this axis, i.e. $\angle K I E=\angle I E K=\angle B E K=45^{\circ}$. So $90^{\circ}$ is a possible value for $\angle B A C$, too. If, on the other hand, $\angle B E K=45^{\circ}$ then $\angle B E K=\angle I E K=\angle I F K=45^{\circ}$. Then  - either $F=E$, which makes the angle bisector $B I$ be an altitude, i.e., which makes triangle $A B C$ isosceles with base $A C$ and hence equilateral and so $\angle B A C=60^{\circ}$, - or $E$ lies between $F$ and $C$, which makes the points $K, E, F$ and $I$ concyclic, so $45^{\circ}=$ $\angle K F C=\angle K F E=\angle K I E=\angle C B I+\angle I C B=2 \cdot \angle I C B=90^{\circ}-\frac{1}{2} \angle B A C$, and so $\angle B A C=90^{\circ}$, - or $F$ lies between $E$ and $C$, then again, $K, E, F$ and $I$ are concyclic, so $45^{\circ}=\angle K F C=$ $180^{\circ}-\angle K F E=\angle K I E$, which yields the same result $\angle B A C=90^{\circ}$. (However, for $\angle B A C=90^{\circ} E$ lies, in fact, between $F$ and $C$, see Figure 2. So this case does not occur.) This proves $90^{\circ}$ and $60^{\circ}$ to be the only possible values for $\angle B A C$.