[ Two tangents drawn from one point
[ $\left.\quad \begin{array}{l}\text { Similar auxiliary triangles }\end{array}\right]$
[ Properties and characteristics of isosceles triangles. $]$
[ Trapezoids (other).

In trapezoid $A B C D$ with bases $A D$ and $B C$, the lateral side $A B$ is equal to 2. The bisector of angle $B A D$ intersects line $B C$ at point $E$. A circle is inscribed in triangle $A B E$, touching side $A B$ at point $M$ and side $B E$ at point $H, M H=1$. Find the angle $B A D$.

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Use the similarity of triangles $B M H$ and $B A E$.

## Solution

Since $B E \| A D$, then $\angle A E B=\angle E A D=\angle B A E$. Therefore, triangle $A B E$ is isosceles.

Consequently, $B E=A B=2$ and $M H \| A E$.

Let $B M=B H=x$. Suppose $K$ is the point of tangency of the given circle with side $A E$. Then $A K=A M=2-$ $x, A E=4-2 x$.

From the similarity of isosceles triangles $M B H$ and $A B E$, it follows that $M H: A E=B M: A B$, or $1 / 4-2 x=x / 2$, or $x^{2}-2 x+1=0$.

From this, $x=1$, meaning these triangles are equilateral. Therefore, $\angle B A D=2 \angle B A E=120^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_a7c46f4bbe36942e4ae1g-03.jpg?height=280&width=644&top_left_y=2101&top_left_x=707)

## Answer

$120^{\circ}$.

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