In △ABC, AB = AC. D, E and F are mid-points of the sides BC, CA and AB respectively. Show that :

(i) AD is perpendicular to FE.

(ii) AD and FE bisect each other.

(i) Given: In △ABC, AB = AC. D, E and F are mid-points of the sides BC, CA and AB.

To prove: AD is perpendicular to FE.

Construction: Draw line AD which meets BC at D and join F and E. M is the intersection point of AD and FE. Join ED and FD.

Proof: In △ABC,

AB = AC

⇒ $\dfrac{1}{2}$ AB = $\dfrac{1}{2}$ AC

⇒ AF = AE ………………(1)

E is mid point of AC and F is mid point of AB. By mid-point theorem,

EF ∥ BC and EF = $\dfrac{1}{2}$ BC

or, EF ∥ BD and EF = BD

Similarly, E is mid point of AC and D is mid point of BC. By mid-point theorem,

ED ∥ AB and ED = $\dfrac{1}{2}$ AB

or, ED ∥ AF and ED = AF

D is mid point of BC and F is mid point of AB. By mid-point theorem,

DF ∥ AC and DF = $\dfrac{1}{2}$ AC

or, DF ∥ AE and DF = AE

From the midpoint theorem, the quadrilateral AEDF has opposite sides equal and parallel, thus it forms a parallelogram.

Since AE = AF, this parallelogram is a rhombus because all sides are equal.

In a rhombus, the diagonals bisect each other at right angles (90°).

Therefore, AD and EF bisect each other at 90°.

Hence, AD is perpendicular to FE.

(ii) Since AEDF is a rhombus (as shown in part (i)), we know that the diagonals of a rhombus bisect each other.

Thus, AD and FE bisect each other at their point of intersection(M).

Hence, AD and FE bisect each other.