Exercise 97

Triangle $ABC$ has $AB=13$, $BC=14$, and $CA=15$. Let $I$ denote the intersection of the internal angle bisectors of $\triangle ABC$. What is $BI$?

Solution

$s=\frac{13+14+15}2=21$, $[ABC]=\sqrt{21\cdot8\cdot7\cdot6}=84$, so $r=\frac{84}{21}=4$.
$BD=s-AC=21-15=6$ (tangent length from $B$).
$BI=\sqrt{BD^2+r^2}=\sqrt{36+16}=\sqrt{52}=\boxed{2\sqrt{13}}$.