12.431 The angle between the planes of two equal right triangles $A B C$ and $A D C$ with a common hypotenuse $A C$ is $\alpha$. The angle between the equal legs $\boldsymbol{A} \boldsymbol{B}$ and $A D$ is $\boldsymbol{\beta}$. Find the angle between the legs $B C$ and $C D$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-306.jpg?height=297&width=388&top_left_y=1191&top_left_x=119)

Drop perpendiculars from vertex $B$ and vertex $D$ to $AC$: they will intersect at one point $M$, since $\triangle ABC = \triangle ADC$. Let $K$ be the midpoint of $BD$.

Since $\triangle ABD, \triangle BMD, \triangle BCD$ are isosceles, $AK, MK$, and $CK$ are perpendicular to $BD$.

Let $\angle BCK = x$.

Then $CB = \frac{BK}{\sin x}, AB = \frac{BK}{\sin \frac{\beta}{2}}, MB = \frac{BK}{\sin \frac{\alpha}{2}}, AC^2 = AB^2 + BC^2$.

But $AB \cdot BC = AC \cdot MB \Rightarrow$

$$
\Rightarrow \frac{BK^2}{\sin x \cdot \sin \frac{\beta}{2}} = \frac{BK^2 \cdot \sqrt{\sin^2 x + \sin^2 \frac{\beta}{2}}}{\sin \frac{\alpha}{2} \cdot \sin x \cdot \sin \frac{\beta}{2}} \Rightarrow \sin^2 x = \sin \frac{\alpha - \beta}{2} \cdot \sin \frac{\alpha + \beta}{2}
$$

From this, we get the answer.

Answer: $\angle BCD = 2x = 2 \arcsin \sqrt{\sin \frac{\alpha + \beta}{2} \cdot \sin \frac{\alpha - \beta}{2}}$.