prove that if the diagonal of a parallelogram bisect each other at right angle, it is a rhombus
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Take quadrilateral ABCD , AC and BD are diagonals which intersect at O.
In △AOB and △AOD
DO = OB ∣ O is the midpoint
AO=AO ∣ Common side
∠AOB = ∠AOD ∣ Right angle
So, △AOB ≅△AOD
So, AB = AD
Similarly, AB = BC = CD = AD can be proved which means that ABCD is a rhombus.
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