Proof . – Let ABCD be a rhombus whose diagonal AC and BD intersect at the point O.
We know that the diagonals of a parallelogram bisect each other.
Also we know that every rhombus is a parallelogram.
Therefore OA=OC and OB=OD.
From triangle(COB) and triangle (COD), we have:
CB=CD sides of rhombus.
Therefore tri(COB)~tri(COD) by SSS congruence.
But <COB +<COD =2 right angles (linear pair)
Thus, <COB=<COD=1 right angles
Hence, the diagonals of a rhombus bisect each other at right angles.
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