If the diagonal of a quadrilateral are equal and bisect each other not at right angle. Prove that the quadrilateral is a rectangle?
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Answer:
Step-by-step explanation:
Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of
its interior angles is 900.
In ΔABC and ΔDCB,
AB = DC (Opposite sides of a parallelogram are equal)
BC = BC (Common)
AC = DB (Given)
By SSS congruence rule,
ΔABC ≅ ΔDCB
So, ∠ABC = ∠DCB
It is known that the sum of measures of angles on the same side of traversal is 1800
∠ABC + ∠DCB = 1800 [AB || CD]
=> ∠ABC + ∠ABC = 1800
=> 2∠ABC = 1800
=> ∠ABC = 900
Since ABCD is a parallelogram and one of its interior angles is 900, ABCD is a rectangle.
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