Question

AC and BD are the diagonalso of kite ABCD with AB=AD while AG and ME are the diagonals of rectangle AEGM. If AD=DG and BG is aline segment then x-y equals
MatHSBC

Answer 1

Answer is 48°

See the photo ,to know how.

Answer 2

We know that ABCD is a kite

So, AB=AD and BC=CD.

And EFGH are mid points of AB ,BC,CD,AD

Now in order to prove that  EFGH is a rectangle

We will first of all construct this:

join AC and BD.

In ΔABD  the E and F are mid points.

so we can write this as:

EF ║BD

EF = 1/2 BD    (this will be an equation 1) (here will be using the mid point theorem)

In Δ BCD

G and H are mid points.

Which means GH║ BD

GH=1/2 BD  (this will be an equation 2)

From the equation (1) and (2)

We can easily see that

EF║ GH

and

EF = GH ( these are the opposite sides of quadrilateral

EFGH is a parallelogram)

We know that  ABCD is a kite so the diagonal intersect at 90.

So ∠AOd= 90

90 is the final answer for this question.

If there is any confusion please leave a comment below.