5. In an acute-angled triangle $\mathrm{ABC}$, a line $s$ passing through vertex A and the median from vertex B divide the height dropped from vertex C into three equal parts. In what ratio does the line $s$ divide the side BC?
5. Answer: 1:4. Since the line l and the median divide the height into three equal parts, the height is divided by the intersection point with the median in the ratio $2: 1$, counting from its vertex, meaning the height from vertex C is also the median of the isosceles triangle $\mathrm{ABC}(\mathrm{AC}=\mathrm{BC})$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1a3173b4782c774e83f4g-1.jpg?height=457&width=460&top_left_y=2056&top_left_x=364)

Let the line s intersect the height CH at point $\mathrm{K}$, and side $\mathrm{BC}$ at point $\mathrm{P}$. Note that triangle AKB is also isosceles. Mark point M on side BC such that KM is parallel to $\mathrm{AB}$.

Angle PKB is equal to the sum of angles KAB and KBA. Since KM is parallel to AB, KM is the angle bisector of angle PKB. Triangles CKM and CHB are similar with a ratio of 3, so $\mathrm{K}=\mathrm{AB} / 6$. Therefore, triangles

PKM and PAB are similar with a ratio of 6, meaning PK:BK $=1: 5$, i.e., $\mathrm{PM}: \mathrm{MB}=1: 5$. Let $\mathrm{x}=\mathrm{CP}, \mathrm{y}=\mathrm{PM}$. Then on one hand $\mathrm{BM}=2 *(\mathrm{x}+\mathrm{y})$, and on the other $\mathrm{BM}=5 \mathrm{y}$. Hence $2 \mathrm{x}=3 \mathrm{y}$, and therefore $\mathrm{CP}: \mathrm{PB}=\mathrm{x} /(6 \mathrm{y})=$ $(3 / 2) / 6=1: 4$.

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