Bikin Yu.A.

In trapezoid $ABCD$, sides $AD$ and $BC$ are parallel, and $AB = BC = BD$. The height $BK$ intersects diagonal $AC$ at point $M$. Find $\angle CDM$.

#
From the equality of sides $AB$ and $BC$ and the parallelism of $BC$ and $AD$, it follows that $AC$ is the bisector of angle $BAD$ (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_a69fc833ac69b1da7555g-08.jpg?height=395&width=766&top_left_y=1969&top_left_x=655)

Since triangle $ABD$ is isosceles, $DM$ is the bisector of angle $ADB$. Let $\angle BAD = \angle BDA = \angle DBC = 2\alpha$.

We can reason in different ways:

First method. From the isosceles triangle $BCD$, we get that $\angle CDB = 90^\circ - \alpha$. Then $\angle CDM = \angle CDB + \angle MDB = 90^\circ$.

Second method. Since $\angle BDM = \angle BAM = \angle BCM = \alpha$, quadrilateral $BCDM$ is cyclic and $\angle CDM = 180^\circ - \angle MBC = 90^\circ$.

Third method. Similarly to the beginning of the solution, we can show that $DC$ is the bisector of the external angle at $D$. The bisectors of the internal and external angles are perpendicular.

[ Inequalities for angles of a triangle ]
[ The larger angle lies opposite the longer side ]
Problem