In a triangle $A B C$, the bisector of $\angle B A C$ meets the side $B C$ at the point $D$. Knowing that $|B D| \cdot|C D|=|A D|^{2}$ and $\angle A D B=45^{\circ}$, determine the angles of triangle $A B C$.
Answer: $\angle B A C=60^{\circ}, \angle A B C=105^{\circ}$ and $\angle A C B=15^{\circ}$.

Suppose the line $A D$ meets the circumcircle of triangle $A B C$ at $A$ and $E$ (see Figure 5). Let $M$ be the midpoint of $B C$ and $O$ the circumcentre of triangle $A B C$. Since the arcs $B E$ and $E C$ are equal, then the points $O$, $M, E$ are collinear and $O E$ is perpendicular to $B C$. From the equality
$\angle C D E=\angle A D B=45^{\circ}$ it follows that $\angle A E O=45^{\circ}$. Since $|A O|=|E O|$, we have $\angle A O E=90^{\circ}$ and $A O \| D M$.

From the equality $|B D| \cdot|C D|=|A D|^{2}$ we obtain $|A D|=|D E|$, which implies that $|O M|=|M E|$. Therefore $|B O|=|B E|$ and also $|B O|=|E O|$. Hence the triangle $B O E$ is equilateral. This gives $\angle B A E=30^{\circ}$, so $\angle B A C=60^{\circ}$. Summing up the angles of the triangle $A B D$ we obtain $\angle A B C=105^{\circ}$ and from this $\angle A C B=15^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_2e42d82bdb4a68411fc6g-08.jpg?height=374&width=458&top_left_y=515&top_left_x=316)

Figure 5