1. Sei $A B C$ ein spitzwinkliges Dreieck mit $A B \neq B C$ und Umkreis $k$. Seien $P$ und $Q$ die Schnittpunkte von $k$ mit der Winkelhalbierenden beziehungsweise der Aussenwinkelhalbierenden von $\angle C B A$. Sei $D$ der Schnittpunkt von $A C$ und $P Q$. Bestimme das Verhältnis $A D: D C$.

## 1st solution:
By the inscribed angle theorem and the fact that $B P$ is the bisector of $\angle A B C, $\angle P C A=$ $\angle P B A=\angle P B C=\angle P A C$, so $\triangle A P C$ is isosceles. Similarly, let $R$ be a point on the line $A B$ such that $B$ lies between $A$ and $R$. Then $\angle Q A C=\angle Q B C=\angle Q B R=180^{\circ}-\angle Q B A=\angle Q C A$, where we've used the fact that $Q B$ is the exterior bisector of $\angle A B C$ and the fact that $A B Q C$ is an inscribed quadrilateral. So $\triangle A Q C$ is also an isosceles triangle. Since $P$ and $Q$ are both at equal distances from $A$ and $C$, $P Q$ is the perpendicular bisector of $A C$, which means that $D$ is the midpoint of the segment $A C$. We therefore obtain that $\frac{A D}{D C}=1$.

## 2nd solution:

As above, we demonstrate that $\triangle A Q C$ is isosceles at $Q$. Furthermore, $\angle C Q P=\angle C B P=$ $\angle A B P=\angle A Q P$, where we use the inscribed angle theorem and the fact that $B P$ is the interior bisector of $\angle A B C$. So $Q D$ is the bisector of $\angle A Q C$ and, since the triangle is isosceles, $Q D$ is the perpendicular bisector of $A B$. We therefore conclude that $\frac{A D}{D C}=1$.

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