## Task 2 - 340712

For a triangle $A B C$, it is required: The angle bisector through $A$ and the perpendicular bisector of $A B$ intersect at a point $D$, which lies on the side $B C$.

a) What size must the angle $\angle A C B$ have in a triangle that meets this requirement and in which the angle $\angle A B C$ has a size of $35^{\circ}$? Draw such a triangle!

b) Draw a triangle that also meets the above requirement and in which the angle $\angle A B C$ has a size of $50^{\circ}$!

c) Is there a triangle that also meets the above requirement and in which the angle $\angle A B C$ has a size of $60^{\circ}$?
a) For the interior angle sizes $\alpha=\angle B A C, \beta=\angle A B C, \gamma=\angle A C B$, it holds that: Since $D$ is a point on the perpendicular bisector of $A B$, it follows that $D A=D B$. By the base angle theorem for triangle $A B D$, we have $\angle B A D=\angle A B D=\beta=35^{\circ}$.

Since $A D$ bisects the interior angle at $A$, it follows that $\alpha=2 \cdot \angle B A D=2 \beta=70^{\circ}$. Therefore, from the angle sum theorem for triangle $A B C$, we get $\gamma=180^{\circ}-(\alpha+\beta)=75^{\circ}$.

b) The right image shows a sought triangle.

c) If there were a triangle of the mentioned type, then as in a), it would follow that $\alpha=2 \beta=120^{\circ}$.

Thus, $\alpha+\beta+\gamma>\alpha+\beta=120^{\circ}+60^{\circ}=180^{\circ}$, which contradicts the angle sum theorem.