ABCD is kite having AB=AD and BC=CD.prove that the figure formed by joining the midpoints of the sides in order is a rectangle
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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.
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