Sides of the triangle from the bisector

Text

In triangle $A\hat{C}B$, right-angled at $\hat{C}$, draw the bisector of angle $\hat{A}$, which meets leg $BC$ at $K$. You know that $\overline{BK}=3$ and $\overline{KC}=2$. Determine all the sides of the triangle.

Solution

Leg $BC=BK+KC=3+2=5$. By the angle bisector theorem the bisector of $\hat A$ divides the opposite side $BC$ into parts proportional to the adjacent sides: $\frac{\overline{BK}}{\overline{KC}}=\frac{\overline{AB}}{\overline{AC}}=\frac32.$ Let $\overline{AC}=2k$ and $\overline{AB}=3k$. Since the triangle is right-angled at $C$, by Pythagoras $\overline{AB}^2=\overline{AC}^2+\overline{BC}^2$: $(3k)^2=(2k)^2+5^2\ \Longrightarrow\ 5k^2=25\ \Longrightarrow\ k=\sqrt5.$ Hence $\boxed{\overline{BC}=5,\qquad \overline{AC}=2\sqrt5\approx4{,}47,\qquad \overline{AB}=3\sqrt5\approx6{,}71.}$