Geometry Triangle Problem

Question

Does anyone have a way to do the below without the Angle Bisector Theorem?

In △ABC, ∠ABC is 90°. Line segments AB and AC have lengths 6 and 10 respectively. D is a point on BC such that AD bisects ∠CAB. Let E be the foot of the perpendicular from D onto AC. What is the perimeter of △CDE?

Thanks

Answer 1

In this triangle, using pythagorus theorem ,we get $BC=8$

$\angle BAD=\angle DAE $ and $\angle BDA=\angle ADE$ and $AD$ is common.

by using ASA congruence criterion $\triangle ABD \cong \triangle AED$

so, $ BD=DE$

perimeter of $\triangle DCE = DE+EC+DC= (BD+DC)+(AC-AE)= 8+ 10-6 = 12 units$