$\underline{\text { Folklore }}$

In a convex quadrilateral $ABCD$, $\angle ABC=90^{\circ}$, $\angle BAC=\angle CAD$, $AC=AD$, and $DH$ is the altitude of triangle $ACD$. In what ratio does the line $BH$ divide the segment $CD$?
Since angle $B A C$ is acute, angle $C A D$ is also acute, so triangle $A C D$ is acute-angled. Therefore, $H$ is an interior point of segment $A C$.

Let $N$ be the intersection point of line $B H$ and segment $C D$ (see figure). Since triangles $A B C$ and $A H D$ are congruent (by hypotenuse and acute angle), $A B = A H$, so triangle $A B H$ is isosceles. Let $\angle H B A = \alpha$, then $\angle C H N = \angle B H A = \alpha$ and $\angle A C N = \frac{1}{2}\left(180^{\circ} - \angle C A D\right) = \frac{1}{2}\left(180^{\circ} - \angle C A B\right) = \alpha$, $\angle N H D = 90^{\circ} - \alpha = \angle N D H$. Thus, $C N = H N = D N$, so $H N$ is the median of triangle $C H D$.

![](https://cdn.mathpix.com/cropped/2024_05_06_95d66af6ee6f6b4b6717g-37.jpg?height=761&width=780&top_left_y=14&top_left_x=639)

## Answer

$1: 1$.