ABC is an isosceles triangle with AB = AC. D, E and F are mid-points of the sides BC, AB and AC respectively. Prove that line segment AD is perpendicular to EF and is bisected by it.

From figure,

In △ABD and △ACD,

△ABC is an isosceles triangle

∴ ∠ABD = ∠ACD

Here D is the mid-point of BC

BD = CD

It is given that AB = AC

∴ △ABD ≅ △ACD (By SAS axiom of congruency)

⇒ ∠ADB = ∠ADC (By c.p.c.t.c)

From figure,

⇒ ∠ADB + ∠ADC = 180°
⇒ ∠ADB + ∠ADB = 180°
⇒ 2∠ADB = 180°
⇒ ∠ADB = 90°.

So, AD is perpendicular to BC.

D and E are mid-points of BC and AB,

By midpoint theorem,

DE || AC or,

DE || AF …….(i)

D and F are mid-points of BC and AC,

By midpoint theorem,

DF || AB or,

DF || AE …….(ii)

Using (i) and (ii) we get,

AEDF is a parallelogram.

Diagonals of parallelogram bisect each other

AD and EF bisect each other.

Since, E and F are mid-points of AB and AC,

By midpoint theorem,

EF || BC

Since, AD is perpendicular to BC and EF || BC.

∴ AD ⊥ EF.

Hence, proved that line segment AD is perpendicular to EF and is bisected by it.