Problem 9.8. On the side $CD$ of trapezoid $ABCD (AD \| BC)$, a point $M$ is marked. A perpendicular $AH$ is dropped from vertex $A$ to segment $BM$. It turns out that $AD = HD$. Find the length of segment $AD$, given that $BC = 16$, $CM = 8$, and $MD = 9$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8cecc131629e5ae42e92g-35.jpg?height=444&width=589&top_left_y=743&top_left_x=432)
Answer: 18.

Solution. Let the lines $B M$ and $A D$ intersect at point $K$ (Fig. 5). Since $B C \| A D$, triangles $B C M$ and $K D M$ are similar by angles, from which we obtain $D K = B C \cdot \frac{D M}{C M} = 16 \cdot \frac{9}{8} = 18$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8cecc131629e5ae42e92g-35.jpg?height=317&width=699&top_left_y=1424&top_left_x=381)

Fig. 5: to the solution of problem 9.8

In the isosceles triangle $A D H$, draw the height and median $D S$. Then, in triangle $A H K$, the segment $D S$ passes through the midpoint of side $A H$ and is parallel to $H K$. Therefore, $D S$ is the midline of this triangle, and $A D = D K = 18$.

Remark. If in a right triangle a point on the hypotenuse is equidistant from two vertices of the triangle, then it is equidistant from all three vertices of the triangle.

## 10th Grade