In the figure given below, ABCD is a kite in which BC = CD, AB = AD. E, F, G are mid-points of CD, BC and AB respectively. Prove that :

(i) ∠EFG = 90°

(ii) The line drawn through G and parallel to FE bisects DA.

Construction,

1. Join AC and BD

2. AC and BD intersect at O.

3. Join EF and FG.

(i) We know that,

Diagonals of a kite intersect at right angles.

∠MON = 90° …….(i)

In △BCD,

E and F are mid-points of CD and BC,

EF || DB and EF = $\dfrac{1}{2}$DB ……(ii)

Since, EF || DB we can say that,

MF || ON.

As sum of opposite angles of a quadrilateral = 180°

∠MON + ∠MFN = 180°

90° + ∠MFN = 180°

∠MFN = 90°.

From figure,

∠EFG = ∠MFN = 90°.

Hence, proved that ∠EFG = 90°.

(ii) From part (i) we get,

FE || BD

Here line through G (GH) is parallel to FE.

∴ GH || FE

or GH || BD.

In △ABD,

GH || BD and G is midpoint of AB,

∴ H is mid-point of AD (By converse of mid-point theorem).

Hence, proved that the line drawn through G and parallel to FE bisects DA.