## Task 5 - 280835

Let $ABC$ be a triangle, $\alpha$ be the measure of the angle $\angle BAC$, and $\beta$ be the measure of the angle $\angle ABC$. The incircle of the triangle touches the side $AB$ at $D$, the side $BC$ at $E$, and the side $AC$ at $F$.

Determine the measure of the angle $\angle FDE$ in terms of $\alpha$ and $\beta$!

Hint: The incircle of a triangle is the circle that touches all three sides of the triangle from the inside.
If the center of the incircle is denoted by $M$, then $DM$ and $FM$ are radii. The lines through $A$ and $B$ and through $A$ and $C$ are tangents to the incircle. Since the tangent and the radius at the point of tangency are perpendicular to each other, we have

$$
\angle ADM = \angle AFM = 90^{\circ}
$$

According to the theorem on the sum of the interior angles of a quadrilateral, for the quadrilateral $ADMF$ we have:

$$
\angle BAC + \angle ADM + \angle DMF + \angle AFM = 360^{\circ}
$$

Using (1) and by the definition of $\alpha$, it follows that

$$
\angle DMF = 180^{\circ} - \alpha
$$

By analogous considerations in the quadrilateral $DBEM$, we obtain

$$
\angle DME = 180^{\circ} - \beta
$$

For the measure of the desired angle $\angle FED$, we have

$$
\angle FDE = \angle MDF + \angle MDE
$$

The angle $\angle MDF$ is the base angle of the isosceles triangle $DMF$; from (2) it follows that

$$
\angle MDF = \frac{180^{\circ} - (180^{\circ} - \alpha)}{2} = \frac{\alpha}{2}
$$

Similarly, from (3) we get: $\angle MDE = \frac{\beta}{2}$. Therefore, from (4) we obtain: $\angle FDE = \frac{\alpha + \beta}{2}$.