# Problem 5. (3 points)

In an isosceles trapezoid $A B C D$, the bisectors of angles $B$ and $C$ intersect on the base $A D$. $A B=50, B C=128$. Find the area of the trapezoid.
Answer: 5472

## Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_06917b52847f1da31476g-2.jpg?height=305&width=600&top_left_y=1955&top_left_x=731)

Let $K$ be the point of intersection of the angle bisectors. Angles $\angle B=\angle C$ are equal as the base angles of an isosceles trapezoid, so their halves are also equal, that is, $\angle K B A=\angle K B C=\angle K C B=\angle K C D$. Additionally, $\angle K B C=\angle A K B$ and $\angle K C B=\angle D K C$ as alternate interior angles.

Thus, triangles $A K B$ and $K C D$ are isosceles, and $A D=A K+K D=A B+C D=2 A B$.

Moreover, triangles $A K B$ and $K B C$ are similar, so $B K^{2}=A B \cdot B C$. We can calculate the height of the isosceles triangle $K B C$ using the Pythagorean theorem:

$$
K H^{2}=B K^{2}-B H^{2}=A B \cdot B C-\frac{B C^{2}}{4}
$$

from which the area of the trapezoid is $\frac{1}{2}(2 A B+B C) \sqrt{A B \cdot B C-\frac{B C^{2}}{4}}$.

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