Question

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.​

Answer 1

(i) It is given that

ABCD is a kite in which BC=CD, AB=AD

E, F, G are midpoints of CD, BC and AB

To prove :

∠EFG=90°

Construction :

Join AC and BD

Construct GH through G paralled to FE

Proof:

We know that,

Diagonals of a kite intersect at right angles

∴∠MON=90°...(1)

In △BCD

E and F are midpoints of CD and BC

⇒EF∥DB and EF=1/2DB...(2) [ By basic proportionality theorem ]

Now, EF∥DB⇒MF∥ON

Similarly, FG∥CA⇒FN∥MO

Therefore, in quadrilateral MFNO,

MF∥ON, FN∥MO  and  ∠MON=90°

⇒MFNO is a square.

∴∠EFG=90°

Your 1st part is given in text and 2nd part is in image.

hope it helps you

Answer 2

Refer the above attachment for the solution!