## Task 6 - 140736

Claudia tells her friend Sabine that she has drawn a triangle $A B C$ in which the altitude from $A$ to $B C$ passes exactly through the intersection of the perpendicular bisector of $A B$ and the angle bisector of $\angle A B C$.

Sabine claims that from this information alone, one can determine the size of the angle $\angle A B C$ without seeing the drawing.

Investigate whether Sabine's claim is correct! If this is the case, determine the size of $\angle A B C$!
Let $S$ be the mentioned intersection point, $E$ the midpoint of side $AB$, and $F$ the foot of the perpendicular from $B$ to $BC$.

Since the line through $B$ and $S$ bisects the angle $\angle ABC$, it follows that $\angle FBS = \angle SBA$. (1)

Since $S$ also lies on the perpendicular bisector of $AB$, $A$, $B$, and $S$ are the vertices of an isosceles triangle with $AS = BS$, and therefore $\angle SAB = \angle SBA$. (2)

From (1) and (2), it follows that $\angle FBS = \angle SAB = \angle SBA$. (3)

Furthermore, $A$, $B$, and $F$ are the vertices of a right triangle. The sum of the sizes of the angles mentioned in (3) is thus $90^\circ$ according to the angle sum theorem. Each of them therefore has a size of $30^\circ$. Sabine's claim is thus correct. The size of the angle $\angle ABC$ is $60^\circ$.

Solutions of the III. Round 1974 taken from [5]

### 4.17 XV. Olympiad 1975

### 4.17.1 I. Round 1975, Class 7