Two circles intersect at points $A$ and $B$. A line through point $B$ intersects the circles at points $C$ and $D$, which lie on opposite sides of line $A B$. The tangents to these circles at points $C$ and $D$ intersect at point $E$. Find $A E$, if $A B=10, A C=16, A D=15$.
Prove that a circle can be circumscribed around quadrilateral $ABCD$, and also that triangles $ABC$ and $ADE$ are similar.

## Solution

From the theorem about the angle between a tangent and a chord, it follows that $\angle BAC = \angle ECD$ and $\angle BAD = \angle EDC$. Since ray $AB$ lies between the sides of angle $CAD$, then

$\angle CAD = \angle BAC + \angle BAD = \angle ECD + \angle EDC = 180^\circ - \angle CED$. Therefore, a circle can be circumscribed around quadrilateral $ADEC$.

Therefore, $\angle BAC = \angle ECD = \angle EAD$, $\angle ACB = \angle AED$. Hence, triangles $ABC$ and $ADE$ are similar by two angles.

Thus, $AE = AD \cdot \frac{AC}{AB} = 24$.

![](https://cdn.mathpix.com/cropped/2024_05_06_476f7e28c93a861c2173g-05.jpg?height=863&width=941&top_left_y=1458&top_left_x=570)

## Answer

24.

| Problem 102426 Topics | Intersecting Circles | Difficulty: $4-$  Grades: 8,9  l. |
| :---: | :---: | :---: |
|  | $\left[\begin{array}{l}{[\text { Auxiliary similar triangles }} \\ {[\text { angles subtending equal arcs and equal chords] }}\end{array}\right]$ |  |
|  | Angle between a tangent and a chord |  |
|  | Inscribed quadrilaterals (other) |  |

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