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

You may use coordinates. Taking AB and AC as x,y axes – Love Invariants Jun 01 '18 at 10:52 — Love Invariants, Jun 01 '18 at 10:52
Have you tried drawing a diagram? – lioness99a Jun 01 '18 at 11:02 — lioness99a, Jun 01 '18 at 11:02

score 2 · Accepted Answer · answered Jun 01 '18 at 14:55

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

Geometry Triangle Problem

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