Berroov S.l.

In a convex quadrilateral $A B C D$, angles $B$ and $D$ are equal, $C D=4 B C$, and the bisector of angle $A$ passes through the midpoint of side $C D$.

What can the ratio $A D: A B$ be?
Let $M$ be the midpoint of side $CD$. Consider point $K$ on ray $AB$, which is symmetric to point $D$ with respect to line $AM$. Since

$\angle ABC = \angle ADM = \angle AKM$, then $BC \| KM$ and point $K$ lies on segment $AB$. Since $CM = DM = KM$, then $\angle DKC = 90^{\circ}$ and $KC \| AM$. Therefore, the sides of triangles $AKM$ and $KBC$ are respectively parallel, so they are similar with a ratio $k = KM / BC = 2$, from which $AD = AK = 2 KB$ and

$$
AD: AB = 2: 3
$$

## Answer

$2: 3$