4.21. The side, bisector, and height of a triangle, all emanating from the same vertex, are equal to 5, 5, and \(2 \sqrt{6}\) cm, respectively. Find the other two sides of the triangle.
4.21. We find (Fig. 4.18) $A H=H D=\sqrt{25-24}=1$ (cm), i.e., $A D=2$ cm. Further, we have $\frac{B C}{D C}=\frac{A B}{A D}=\frac{5}{2}$. It is known that the square of the length of the angle bisector is expressed by the formula $B D^{2}=A B \cdot B C - A D \cdot D C$ (see formula (1.37)). Thus, $25=5 \cdot 5 x - 2 \cdot 2 x$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-582.jpg?height=401&width=508&top_left_y=784&top_left_x=644)

Fig. 4.18 or $25=21 x$, from which $x=\frac{25}{21}$. Therefore, $A C=2+\frac{50}{21}=4 \frac{8}{21}(\mathrm{~cm}), B C=\frac{5 \cdot 25}{21}=5 \frac{20}{21}(\mathrm{~cm})$.

Answer: $4 \frac{8}{21}$ and $5 \frac{20}{21} \mathrm{~cm}$.