Question

In the given figure ac and bd are the diagonals of a kite abcd with ab=ad while ag and me are the diagonals of a rectangle aegm. if ad=dg and bg is a line segment then x-y equals to

Answer 1

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

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)

Now since ABCD is a kite ,diagonal intersect at 90.

So ∠AOd= 90.