Find the sine of angle $A D C$.

## Solution

Let $A M$ be the perpendicular dropped from point $A$ to side $C D$. The ray $A C$ passes between the sides of angle $B A M$, so

$$
\angle B A M=\angle B A C+\angle C A M .
$$

Let $R$ be the radius of the circumscribed circle of quadrilateral $A B C D$. Then

$$
\begin{aligned}
R & =\frac{A D}{2 \sin \angle A C D}=\frac{7}{\sqrt{3}} \\
\sin \angle B A C & =\frac{B C}{2 R}=\frac{3 \sqrt{3}}{14}, \angle C A M=30^{\circ} .
\end{aligned}
$$

Therefore,

$$
\begin{aligned}
\sin \angle M A D= & \sin \left(\angle B A C+30^{\circ}\right)=\frac{3 \sqrt{3}}{14} \cdot \frac{\sqrt{3}}{2}+\frac{1}{2} \cdot \frac{13}{14}=\frac{11}{14} \\
& \sin \angle A D C=\cos \angle M A D=\frac{5 \sqrt{3}}{14}
\end{aligned}
$$

$$
A C=2 R \sin \angle A D C=\frac{2 \cdot 7}{\sqrt{3}} \cdot \frac{5 \sqrt{3}}{14}=5
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-09.jpg?height=503&width=489&top_left_y=232&top_left_x=790)

## Answer

5.
8,9

In a convex quadrilateral $A B C D$, diagonal $A C$ is drawn, $A D=7, B C=3, \angle A C D=60^{\circ}$. It is known that points $A, B, C, D$ lie on the same circle, and the perpendicular from point $A$ to side $C D$ bisects angle $\angle B A D$. Find the diagonal $A C$.