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