In the drawing below, $ABC$ is a triangle with side lengths $AB=4 \, \text{cm}, AC=$ $6 \, \text{cm}$. Additionally, $\angle DAH=\angle HAB, \angle AHB=90^{\circ}$, and $M$ is the midpoint of $BC$. Find the length of the segment $MH$.

![](https://cdn.mathpix.com/cropped/2024_05_01_12f29789e10e6749ebb9g-25.jpg?height=549&width=465&top_left_y=590&top_left_x=867)
Solution

Since $A H$ is the altitude and bisector of vertex $A$, it follows that triangle $A B D$ is isosceles. Therefore, $H$ is the midpoint of segment $B D$ and $A B=A D$. Consequently, $M H$ is a midline of triangle $C B D$ and thus

$$
\begin{aligned}
M H & =\frac{C D}{2} \\
& =\frac{A C-A D}{2} \\
& =\frac{A C-A B}{2} \\
& =1 \mathrm{~cm} .
\end{aligned}
$$