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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.

User Sylbru
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1 Answer

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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.

User Mabounassif
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