Question

AE is an altitude of an isosceles triangle ABC in which AB = AC. Show that AE bisects BC

Answer 1

Answer:

∆ABC is an isosceles triangle.

so,AB=AC

Also AE is the altitude

so,angle AEC=AEB=90°

to prove=BE=CE

=angle BAE= Angle CAE

in ∆AEB and AEC

angleAEC=angle AEB=90°(both 90°)

AB=AC(from 1)

AE=AE(common)

∆AEB ≈∆AEC (R.H.S congruency)

hence by CPCT

BE=EC & ∆BAC =EAC

Answer 2

CRM

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