Let ABC be an isosceles triangle in which AB=AC. If D, E, F be the mid-points of the sides BC, CA and AB respectively, show that the segment AD and EF bisect each other at right angles.
Given,
ABC is an isosceles triangle , D,E,F are mid points of side BC, CA,AB
∴AB││DF and AC││FD
ABDF is a parallelogram
AF=DE and AE = DF
=
DE=DF (∵AB=AC)
AE=AF=DE=DF
ABDF is a rhombus
= AD and FE bisect each other at right angle.
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