In trapezoid $B C D E$, the base $B E=13$, the base $C D=3, C E=10$. On the circle circumscribed around trapezoid $B C D E$, a point $A$ different from $E$ is taken such that $C A=10$. Find the length of the segment $B A$ and the area of pentagon $A B C D E$.

#
Prove that $\angle A C B=\angle C D E$.

## Solution

Since the trapezoid $B C D E$ is inscribed in a circle, it is isosceles. Note that angle $C D E$ is obtuse, so for any point $X$ different from $E$ and lying on the arc $C E$ containing point $D$, $C X<C E=10$ (in triangle $C D E$, the side opposite the obtuse angle is the largest). Therefore, the point cannot lie on this arc. Point $A$ cannot lie on the arc $B C$ not containing point $D$ ( $C B=D E<C E=10$). Thus, point $A$ lies on the arc $B E$ not containing point $C$. We will prove the equality of angles $\angle A C B$ and $\angle C E D$. Indeed, $\angle B A C=\angle B E C=\angle D C E$, and since

$$
\angle A B E=\angle A C E=\angle A D E \text { and } \angle C B E=\angle C A E=\angle C E A=\angle A D C \text {, }
$$

then

$$
\angle A B C=\angle A B E+\angle C B E=\angle A D E+\angle A D C=\angle C D E .
$$

Therefore, the angles $\angle A C B=\angle C E D$ (as the remaining angles of triangles $A B C$ and $C D E$) are also equal. Consequently, the chords on which these angles subtend are equal, i.e., $A B=C D=3$. Let $\angle A B E=\angle A C E=\alpha$. From triangles $A B E$ and $A C E$ using the cosine theorem, we find that

$$
A E^{2}=A B^{2}+B E^{2}-2 A B \cdot B E \cdot \cos \alpha=9+169-2 \cdot 3 \cdot 13 \cdot \cos \alpha=178-78 \cos \alpha
$$

$$
A E^{2}=A C^{2}+C E^{2}-2 A C \cdot C E \cdot \cos \alpha=100+100-2 \cdot 10 \cdot 10 \cdot \cos \alpha=200-200 \cos \alpha
$$

From the equation $178 - 78 \cos \alpha=200-200 \cos \alpha$, we find $\cos \alpha=\frac{11}{61}$. Then

$$
\sin \alpha=\sqrt{1-\cos ^{2} \alpha}=\sqrt{1-\frac{11^{2}}{61^{2}}}=\frac{\sqrt{(61-11)(61+11)}}{61}=\frac{60}{61}
$$

Therefore,

$$
S_{\triangle \mathrm{ABE}}=\frac{1}{2} A B \cdot B E \cdot \sin \alpha=\frac{1}{2} \cdot 3 \cdot 13 \cdot \frac{60}{61}=\frac{3 \cdot 13 \cdot 30}{61}
$$

Let $C H$ be the height of the isosceles trapezoid $B C D E$. Then

$$
E H=\frac{B E+C D}{2}=\frac{13+3}{2}=8
$$

From the right triangle $C H E$, we find that

$$
C H=\sqrt{C E^{2}-E H^{2}}=\sqrt{100-64}=6 .
$$

Therefore, $S B C D E=\frac{B E+C D}{2} \cdot C H=8 \cdot 6=48$. Consequently,

$$
S B C D E A=S B C D E+S_{\triangle \mathrm{ABE}}=48+\frac{3 \cdot 13 \cdot 30}{61}=\frac{4098}{61}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_f7e07b07084714df1adcg-38.jpg?height=1001&width=1037&top_left_y=1329&top_left_x=522)

Answer

3; $\frac{4098}{61}$.