Answer:
The problem is symmetrical, so proof for ΔCDG can serve as a model for proof for ΔCEH.
Explanation:
∠DGA ≅ ∠DCA ≅ 90° . . . . given
∠GAD ≅ ∠CAD . . . . definition of angle bisector AD
AD ≅ AD . . . . reflexive property
ΔDGA ≅ ΔDCA . . . . AAS congruence theorem
CD ≅ GD . . . . CPCTC
∴ ΔCDG is isosceles . . . . definition of isosceles triangle (2 sides congruent)
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To do the same for ΔCEH, replace "D" with "E", replace "G" with "H", and replace "A" with "B". The rest of the logic applies.