In trapezoid $A B C E$ base $A E$ is equal to $16, C E=8 \sqrt{3}$. The circle passing through points $A, B$ and $C$ intersects line $A E$ again at point $H ; \angle A H B=60^{\circ}$. Find $A C$.
Apply the Law of Cosines.

## Solution

Let $O_{1}$ be the center of the given circle, $N$ be its point of tangency with the line $A C$, and $K$ be with the side $B C$.

From the right triangle $A O_{1} N$, we find that

$$
A N=O_{1} N \operatorname{ctg} 30^{\circ}=\sqrt{3}
$$

On the other hand, $A N$ is equal to the semiperimeter $p$ of triangle $A B C$. The radius of the inscribed circle of triangle $A B C$ is

$$
\frac{S_{\triangle A B C}}{p}=\frac{2 \sqrt{3}-3}{\sqrt{3}}=2-\sqrt{3} \text {. }
$$

If $O$ is the center of this circle, and $P$ is the point of tangency with the side $A C$, then

$$
O_{1} O=A O_{1}-A O=2 O_{1} N-2 O P=2(\sqrt{3}-1)
$$

Let $F$ be the foot of the perpendicular dropped from point $O$ to the extension of segment $O_{1} K$. Then

$$
\cos \angle O O_{1} F=\frac{O_{1} F}{O_{1} O}=\frac{1+2-\sqrt{3}}{2(\sqrt{3}-1)}=\frac{\sqrt{3}}{2}
$$

Therefore, $\angle O O_{1} F=30^{\circ}$. If segments $O O_{1}$ and $B C$ intersect at point $Q$, then

$$
\angle A Q B=\angle O_{1} Q K=60^{\circ}
$$

Thus,

$$
\angle B C A=\angle A Q B-\angle Q A C=60^{\circ}-30^{\circ}=30^{\circ}
$$

## Answer

8. 

The trapezoid $A B C H$ is inscribed in a circle, so it is isosceles. Therefore, $\angle C A H=\angle A H B=60^{\circ}$. Let $A C=x$ and apply the Law of Cosines to triangle $A C E$:

$$
(8 \sqrt{3})^{2}=x^{2}+16^{2}-16 x
$$

From this, we find that $x=8$.