5. The incircle of triangle $ABC$ touches sides $AB$ and $AC$ at points $D$ and $E$ respectively. Point $I_{A}$ is the excenter of the excircle opposite side $BC$ of triangle $ABC$, and points $K$ and $L$ are the midpoints of segments $DI_{A}$ and $EI_{A}$ respectively. Lines $BK$ and $CL$ intersect at point $F$, which lies inside angle $BAC$. Find $\angle BFC$, if $\angle BAC=50^{\circ}$. (The excircle touches side $C$ and the extensions of sides $AB$ and $AC$ beyond points $B$ and $C$ respectively.)
5. The circle inscribed in triangle $ABC$ touches sides $AB$ and $AC$ at points $D$ and $E$ respectively. Point $I_{A}$ is the center of the excircle opposite side $BC$ of triangle $ABC$, and points $K$ and $L$ are the midpoints of segments $DI_{A}$ and $EI_{A}$ respectively. Lines $BK$ and $CL$ intersect at point $F$, which lies inside angle $BAC$. Find $\angle BFC$ if $\angle BAC = 50^{\circ}$. (The excircle touches side $BC$ and the extensions of sides $AB$ and $AC$ beyond points $B$ and $C$ respectively.)

Answer: $130^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1dddad72584c2d20379ag-2.jpg?height=514&width=623&top_left_y=1034&top_left_x=725)

Fig. 7: for the solution of problem 5.

Let $X$ and $Y$ be the points of tangency of the excircle opposite side $BC$ with the extensions of sides $AB$ and $AC$ respectively (Fig. 7). Right triangles $DXI_{A}$ and $EYI_{A}$ are symmetric with respect to the angle bisector of angle $A$, hence they are equal and isosceles triangles $DKX$ and $YLE$. In these triangles, points $B$ and $C$ are marked on the bases such that $BD = CY = p - b$, where $p$ is the semiperimeter of triangle $ABC$, and $b$ is the length of side $AC$. Consequently, we conclude that $\angle KBD = \angle LCY$ and, thus, quadrilateral $ABFC$ is cyclic. This leads to the relation $\angle BFC = 180^{\circ} - \angle BAC = 130^{\circ}$.

## Criteria

+ Correct solution - 7 points.