[ Law of Sines ]

In a convex quadrilateral $A B C D$, the segment $C M$, connecting vertex $C$ with point $M$ located on side $A D$, intersects diagonal $B D$ at point $K$. It is known that $C K: K M = 2: 1, C D: D K = 5: 3$ and $\angle A B D + \angle A C D = 180^{\circ}$. Find the ratio of side $A B$ to diagonal $A C$.
Consider triangles $A B D, A C D, M D K, M C C$ and use the Law of Sines.

## Solution

Let $\angle A D B=\alpha, \angle A D C=\beta$. Consider triangles $A B D$ and $A C D$. By the Law of Sines,

$$
\frac{A B}{\sin \alpha}=\frac{A D}{\sin \angle A B D}=\frac{A D}{\sin \left(180^{\circ}-\angle A C D\right)}=\frac{A D}{\sin \angle A C D}=\frac{A C}{\sin \beta},
$$

from which we find that $\frac{A B}{A C}=\frac{\sin \alpha}{\sin \beta}$.

Let $M K=x, C D=5 y, \angle C M D=\gamma$. Consider triangles $M D K$ and $M D C$. By the Law of Sines,

$$
\frac{x}{\sin \alpha}=\frac{3 y}{\sin \gamma}, \frac{3 x}{\sin \beta}=\frac{5 y}{\sin \gamma}
$$

from which we find that

$$
\sin \alpha=\frac{x \sin \gamma}{3 y}, \sin \beta=\frac{3 x \sin \gamma}{5 y}
$$

Therefore,

$$
\frac{A B}{A C}=\frac{\sin \alpha}{\sin \beta}=\frac{5}{9}
$$

## Answer

$5: 9$.