Finding the length of an angle bisector in a (right) triangle

Question

We have a triangle $\triangle ABC$ such that $AB = 6,$ $BC = 8,$ and $CA = 10$. If $AD$ is an angle bisector such that $D$ is on $BC$, then find the value of $AD^2.$

So far I used the angle bisector theorem to get $CD/DB = AC/AB$ so $CD/DB = 10/6$. What should I do next?

Answer 1

I am attaching the solution as a picture. Please go through it.

Answer 2

For a triangle $ABC$ with side lengths $a=BC,\,b=AC,\,c=AB$ the squared length of the angle bisector through $A$ is $$ l_A^2 = bc\left(1-\frac{a^2}{(b+c)^2}\right) $$ due to Stewart's theorem. In our case $ABC$ is a right triangle ($b^2=a^2+c^2$) hence it is probably faster to directly exploit the bisector theorem and the Pythagorean theorem. One way or another, the answer is $l_A^2=\color{red}{45}$.