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Question
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Solution
Let ABCD be a parallelogram. To show that ABCD is a rectangle, we must prove that one of its interior angles is 90°.
In ΔABC and ΔDCB,
AB = DC ...(Opposite sides of a parallelogram are equal)
BC = BC ...(Common)
AC = DB ...(Given)
∴ ΔABC ≅ ΔDCB ...(By SSS Congruence rule)
⇒ ∠ABC = ∠DCB
It is known that the sum of the measures of angles on the same side of transversal is 180°.
∠ABC + ∠DCB = 180° ...(AB || CD)
⇒ ∠ABC + ∠ABC = 180°
⇒ 2∠ABC = 180°
⇒ ∠ABC = 90°
Since ABCD is a parallelogram and one of its interior angles is 90°, ABCD is a rectangle.
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