In δabc (m∠c = 90°), the points d and e are the points where the angle bisectors of ∠a and ∠b intersect respectively sides bc and ac . point g ∈ ab so that dg ⊥ ab and h ∈ ab so that eh ⊥ ab . prove that δceh and δcdg are isosceles.

The problem is symmetrical, so proof for ΔCDG can serve as a model for proof for ΔCEH.

∠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)

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.

answer: i'm not going to give you the answer, but some hints to work it out.

step-by-step explanation: so, if there is a straight line that means that the line angle is 180°. so we subtract the 2 angle measures provided, (15 and 38) from 180. so the answer is 180 - (38 + 15). (you have to solve the equation)