In an isosceles triangle $A B C$, points $D$ and $E$ divide the lateral sides in the ratio $B D: D A=B E: E C=n$. Find the angles of the triangle if $A E \perp C D$.

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Using the similarity of triangles, find $\operatorname{tg} \angle D A E$.

## Solution

Since triangle $A B C$ is isosceles, then $A D E C$ is an isosceles trapezoid. Let $K$ be the point of intersection of its diagonals. Triangle $D B E$ is similar to triangle $A B C$ (with a coefficient of ${ }^{n} / n+1$), and triangle $D K E$ is similar to triangle $C K A$ (with the same coefficient). Therefore, $\operatorname{tg} \angle K A D=D K / A K=n / n+1, \operatorname{tg} \angle K A C=1$.

Thus, $\operatorname{tg} \angle B A C=\operatorname{tg}(\angle K A D+\angle K A C)=\frac{1+\frac{n}{n+1}}{1-\frac{n}{n+1}}=2 n+1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6ed8f1205731b80a0039g-39.jpg?height=714&width=374&top_left_y=729&top_left_x=842)

## Answer

$\operatorname{arctg}(2 n+1)$