10.240. The length of the base of an isosceles triangle is $12 \mathrm{~cm}$, and the side lengths - $18 \mathrm{cm}$. Heights are drawn to the lateral sides of the triangle. Calculate the length of the segment whose ends coincide with the bases of the heights.
## Solution.

Let in $\triangle A B C$ (Fig. 10.47) $A B=B C=18 \mathrm{~cm}, A C=12 \mathrm{~cm}, A N, C M$, $B D$ - altitudes. $A D=\frac{1}{2} A C=6$ cm. $M N \| A C$ ( $\triangle A B C$ - isosceles), then $\triangle M B N \sim \triangle A B C \cdot \triangle B C D \sim \triangle A N C$ (both are right triangles, $\angle A C N$ - common). Then $\frac{N C}{C D}=\frac{A C}{B C}, N C=\frac{C D \cdot A C}{B C}=4$. Therefore, $B N=B C-N C=$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0740.jpg?height=111&width=1147&top_left_y=1739&top_left_x=95)
Answer: $\frac{28}{3}$ cm.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0741.jpg?height=537&width=507&top_left_y=77&top_left_x=158)

Fig. 10.48

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0741.jpg?height=547&width=507&top_left_y=74&top_left_x=671)

Fig. 10.49