SOLUTION-
Triangle ABC is a right angle at C. The angles bisectors of BAC and ABC meet BC and AC at D and E respectively. The points H and G are the feet of the perpendiculars from E and D to AB.
As AD is an angle bisector,
⇒ ∠CAD = ∠BAD
⇒ ∠CAD = ∠GAD ( ∵ ∠BAD = ∠GAD)
The two right triangles CAD and GAD ( ∵ m∠G = 90° = m∠C ) share a hypotenuse and have equal acute angles,
⇒ ΔCAD = ΔGAD
⇒ CD = DG
∴ ΔCDG is an isosceles triangle. (Proved)
Likewise,
As BE is an angle bisector,
⇒ ∠CBE = ∠ABE
⇒ ∠CBE = ∠HBE ( ∵ ∠ABE = ∠HBE)
The two right triangles CBE and HBE ( ∵ m∠H = 90° = m∠C ) share a hypotenuse and have equal acute angles,
⇒ ΔCBE = ΔHBE
⇒ CE = EH
∴ ΔCEH is an isosceles triangle. (Proved)
