Question

First question please

Answer 1

ABCD is a kite and AB=AD and BC=CD.

i) EFGH are mid points of AB ,BC,CD,AD resp.
To prove that EFGH is a rectangle.

Construction: join AC and BD.
In ΔABD , E and F are mid points.
so we can write ,EF ║BD and
EF = 1/2 BD ---------(1) (using mid point theorem)

Now , in Δ BCD , G and H are mid points.
So, GH║ BD and
GH=1/2 BD ------------>(2) (by mid point th.)
From (1) and (2) EF║ GH and EF = GH ( opposite sides of quad. EFGH are parallel and equal so EFGH is a parallelogram)

ii) Now since ABCD is a kite ,diagonal intersect at 90.
So ∠efg= 90.


I hope it is helped you