4. Given a right triangle $ABC$, such that for the legs $AC$ and $BC$ it holds $\overline{AC}<\overline{BC}$. The lengths of the median and the altitude dropped to the hypotenuse in this triangle are $m$ and $h$, respectively. Calculate the length of the angle bisector of the right angle.
Solution. Since $\triangle ABC$ is a right triangle, it follows that $\overline{CM}=m=\overline{BM}$, i.e., $\triangle BMC$ is isosceles (Figure 1), so $\measuredangle BCM=\beta=\measuredangle ABC$. Clearly, $\measuredangle ACH=\beta$ (from the right triangle $AHC$). Since $\beta<45^{\circ}$ (because $\overline{AC}<\overline{BC}$), it follows that the angle bisector $CL$ of $\measuredangle HCM$ is also the angle bisector of $\measuredangle ACB$, so $\measuredangle HCL=\measuredangle MCL=\phi=45^{\circ}-\beta$.

![](https://cdn.mathpix.com/cropped/2024_06_05_9b969ba4368c118d4aedg-2.jpg?height=363&width=642&top_left_y=924&top_left_x=381)

Figure 1

![](https://cdn.mathpix.com/cropped/2024_06_05_9b969ba4368c118d4aedg-2.jpg?height=409&width=349&top_left_y=920&top_left_x=1056)

Figure 2

Let $x=\overline{HL}$ and $y=\overline{LM}$, and let $F$ be the foot of the perpendicular dropped from $L$ to $CM$ (Figure 2). Since $\triangle CHL \cong \triangle CFL$, we have $\overline{LF}=x$. Furthermore, $\triangle HMC \sim \triangle FML$, so we have: $m:h=y:x$, i.e., $y=\frac{mx}{h}$. Further, from the right triangle $\triangle HCM$ we find:

$$
h^{2}+(x+y)^{2}=m^{2} ; h^{2}+\left(x+\frac{mx}{h}\right)^{2}=m^{2} \text{ i.e. } x^{2}=h^{2} \cdot \frac{m-h}{m+h}
$$

Finally, from the right triangle $\triangle HLC$ we get:

$$
l^{2}=h^{2}+x^{2}=h^{2}+h^{2} \frac{m-h}{m+h}=h^{2} \frac{2m}{m+h} \text{ i.e. } l=h \sqrt{\frac{2m}{m+h}}
$$

## 2nd Year