Problem 8.7.1. Given trapezoid $A B C D (B C \| A D)$. Point $H$ on side $A B$ is such that $\angle D H A=$ $90^{\circ}$. It is known that $C H=C D=13$ and $A D=19$. Find the length of segment $B C$.

![](https://cdn.mathpix.com/cropped/2024_05_06_0c3ddd33c7236a6f213bg-10.jpg?height=412&width=579&top_left_y=309&top_left_x=437)
Answer: 9.5.

Solution. Extend rays $AB$ and $DC$ until they intersect at point $X$ (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_0c3ddd33c7236a6f213bg-10.jpg?height=505&width=505&top_left_y=920&top_left_x=466)

Figure for the solution to problem 8.7.1

In the isosceles triangle $HCD$, we have $\angle CHD = \angle CDH$. In the right triangle $XHD$, we have $\angle HXD = 90^{\circ} - \angle XDH = 90^{\circ} - \angle CHD = \angle XHC$.

Therefore, triangle $XHC$ is isosceles, and $CX = CH = CD$. Then, in triangle $AXD$, segment $BC$ is the midline (since $BC \| AD$, and point $C$ is the midpoint of segment $XD$), so $BC = \frac{1}{2} AD = 9.5$.