Problem — Segments cut by the bisector on the opposite side

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In a triangle $ABC$ the sides measure $\overline{AB}=9$, $\overline{AC}=6$ and $\overline{BC}=10$. The bisector of the angle $\hat A$ meets the side $BC$ at the point $H$. Find the lengths of the two segments $\overline{BH}$ and $\overline{HC}$ into which the side $BC$ is divided.

Solution

By the angle-bisector theorem the bisector of $\hat A$ divides the opposite side $BC$ into parts proportional to the sides $AB$ and $AC$: $\frac{\overline{BH}}{\overline{HC}}=\frac{\overline{AB}}{\overline{AC}}=\frac{9}{6}=\frac{3}{2}.$ Set $\overline{BH}=3t$ and $\overline{HC}=2t$. Since $H$ lies on the segment $BC$, the sum of the two segments is the whole side: $\overline{BH}+\overline{HC}=\overline{BC}\ \Longrightarrow\ 3t+2t=10\ \Longrightarrow\ 5t=10\ \Longrightarrow\ t=2.$ Hence $\overline{BH}=3t=6,\qquad \overline{HC}=2t=4.$ $\boxed{\overline{BH}=6,\qquad \overline{HC}=4.}$