Question

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.

Answer 1

, an isosceles triangle is given with D,E and F as the mid-points of BC, CA and AB respectively as shown below:

We need to prove that the segment AD and EF bisect each other at right angle.

Let’s join DF and DE.

In , D and E are the mid-points of BC and AC respectively.

Theorem states, the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

Therefore, we get: Or

Similarly, we can get

Therefore, AEDF is a parallelogram

We know that opposite sides of a parallelogram are equal.

and

Also, from the theorem above we get

Thus,

Similarly,

It is given that , an isosceles triangle

Thus,

Therefore,

Also,

Then, AEDF is a rhombus.

We know that the diagonals of a rhombus bisect each other at right angle.

Therefore, M is the mid-point of EF and

Hence proved.