In the adjoining figure, ABCD is a kite in which AB = AD and CB = CD. If E, F, G are respectively the mid-points of AB, AD and CD, prove that :

(i) ∠EFG = 90°

(ii) If GH || FE, then H bisects CB.

Join AC and BD, AC and BD intersects at O.

(i) We know that,

Diagonals of a kite intersect at right angles.

∠MON = 90° …(1)

By mid-point theorem,

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

In △ABD,

E and F are mid-points of AB and AD,

EF || BD and EF = $\dfrac{1}{2}$ BD

By converse of mid-point theorem,

The straight line drawn through the mid-point of one side side of a triangle parallel to another, bisects the third side.

In △ABO,

E is the mid-point of AB and EF || BD so, EM || BO,

∴ M is the mid-point of AO

⇒ AM = MO

In △AOD,

M and F are mid-points of AO and AD,

MF || OD and MF = $\dfrac{1}{2}$ OD …(1)

In △ADC,

G and F are mid-points of CD and AD,

FG || AC and FG = $\dfrac{1}{2}$ AC (By mid-point theorem)

In △AOD,

F is the mid-point of AD and FG || AC so, FN || AO,

∴ N is the mid-point of OD (By converse of mid-point theorem)

⇒ ON = ND

From eq.(1), we have:

MF || OD and MF = $\dfrac{1}{2}$ OD

⇒ MF = ON

∴ OMFN is a parallelogram.

We know that,

Opposite angles of a parallelogram are equal.

⇒ ∠MON = ∠MFN = 90°

From figure,

∠EFG = ∠MFN = 90°

Hence, proved that ∠EFG = 90°.

(ii) EF || BD (Proved above)

A line through G is parallel to EF.

∴ GH || FE

or GH || BD

In △BCD,

GH || BD and G is mid-point of CD.

By converse of mid-point theorem,

The straight line drawn through the mid-point of one side side of a triangle parallel to another, bisects the third side.

∴ H is the mid-point of BC (By converse of mid-point theorem)

Hence, proved that if GH || FE, then H bisects CB.