Q13 of 95 Page 15

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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