[ Auxiliary similar triangles

The Law of Sines

In triangle $A B C$, angles $B$ and $C$ are given. The bisector of angle $A$ intersects side $B C$ at point $D$, and the circumcircle of triangle $A B C$ at point $E$.

Find the ratio $A E: D E$.

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Prove that triangle $B D E$ is similar to triangle $A B E$.

## Solution

Since $\angle D B E=\angle C B E=\angle C A E=\angle B A E$, triangle $B D E$ is similar to triangle $A B E$. Therefore, ${ }^{D E} /{ }_{B E}$ $={ }^{B E} /_{A E}, D E={ }^{B E^{2} /_{A E}}$.

Let $\angle A=\alpha, R$ be the radius of the circle. Then $A E=2 R \sin \angle A B E=2 R \sin (\beta+\alpha / 2)=2 R \sin \left(\beta+90^{\circ}-\beta / 2-\gamma / 2\right)=$ $2 R \cos (\gamma / 2-\beta / 2)$

$B E=2 R \sin \angle B A E=2 R \sin \alpha / 2=2 R \cos (\gamma / 2+\beta / 2)$.

Therefore, $\frac{A E}{D E}=\frac{A E^{2}}{B E^{2}}=\frac{\cos ^{2} \frac{\gamma-\beta}{2}}{\cos ^{2} \frac{\gamma+\beta}{2}}=\frac{1+\cos (\gamma-\beta)}{1+\cos (\gamma+\beta)}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ae1b0d3bbbf9aa5c38e8g-44.jpg?height=455&width=546&top_left_y=1779&top_left_x=756)

## Answer

$\frac{1+\cos (\gamma-\beta)}{1+\cos (\gamma+\beta)}$