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The height $A H$ of triangle $A B C$ is equal to its median $B M$. On the extension of side $A B$ beyond point $B$, point $D$ is marked such that $B D=A B$. Find the angle $B C D$.
From point $M$, drop a perpendicular $ME$ to side $BC$. Since $ME \parallel AH$, $ME$ is the midline of triangle $AHC$ (see figure). Therefore, $ME = \frac{1}{2} AH = \frac{1}{2} BM$.

In the right triangle $BME$, the leg $ME$ is half of the hypotenuse $BM$, so $\angle MBE = 30^{\circ}$. Moreover, $BM$ is the midline of triangle $ACD$, so $BM \parallel CD$. Therefore, $\angle BCD = \angle MBC = 30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c1015ea0b2d2805d8876g-22.jpg?height=623&width=669&top_left_y=2024&top_left_x=703)

Answer

$30^{\circ}$.