10-8. In trapezoid $A B C D$, the bases $A D$ and $B C$ are 8 and 18, respectively. It is known that the circumcircle of triangle $A B D$ is tangent to the lines $B C$ and $C D$. Find the perimeter of the trapezoid.
Answer: 56.

Solution. Let's make the following observation. Through point $B$ on the circle, a line passes parallel to the chord $AD$. It is clear that then $B$ is the midpoint of the arc $AD$, that is, $BA = BD$ (indeed, $\angle 1 = \angle 2$ as alternate interior angles, $\angle 1 = \angle 3$ by the theorem on the angle between a tangent and a chord, therefore, $\angle 2 = \angle 3$).

![](https://cdn.mathpix.com/cropped/2024_05_06_1d29d5cb430346a58eb5g-28.jpg?height=551&width=619&top_left_y=2226&top_left_x=724)

Next, from point $C$, two tangents are drawn to the circle; hence, they are equal: $CD = CB = 18$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1d29d5cb430346a58eb5g-29.jpg?height=563&width=1102&top_left_y=444&top_left_x=474)

It remains to find the length of side $AB$, which we will denote as $x$. For this, note that $\angle CBD = \angle BDA$ as alternate interior angles. Therefore, isosceles triangles $CBD$ and $BDA$ are similar as isosceles triangles with equal angles at the base. From the similarity, we get $\frac{x}{8} = \frac{18}{x}$, from which $x = 12$. Thus, the perimeter of the trapezoid is $12 + 18 + 18 + 8 = 56$.

## 11th Grade