10.317. In triangle $A B C$, the following are known: $B C=15 \text{ cm}, A C=14 \text{ cm}$, $A B=13$ cm. Calculate the area of the triangle enclosed between the height and the bisector drawn from vertex $B$.
## Solution.

Let $B H$ and $B K$ be the altitude and the angle bisector of $\triangle A B C$ (Fig. 10.106). Since $15^{2}<13^{2}+14^{2}$, $\triangle A B C$ is an acute triangle and point $H$ lies on the segment $A C$. Using Heron's formula, we find: $S_{\triangle A B C}=84 \text{ cm}^2$. Then $B H=\frac{2 S_{\triangle A B C}}{A C}=12$ cm. Let $A K=x$ cm. Then $C K=(14-x)$ cm and by the property of the angle bisector of a triangle, $\frac{A K}{A B}=\frac{C K}{B C}$, hence $\frac{x}{13}=\frac{14-x}{15}$; $x=6.5$. From $\triangle A H B \quad (\angle A H B=90^{\circ}): A H=\sqrt{A B^{2}-B H^{2}}=5$ cm. Then $H K=A K-A H=1.5 \text{ cm}$. The required area $S_{\triangle B H K}=\frac{1}{2} B H \cdot H K=9 \text{ cm}^2$.

Answer: $9 \text{ cm}^2$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0790.jpg?height=432&width=682&top_left_y=428&top_left_x=335)

Fig. 10.107