Q. 54.8( 8 Votes )
Prove that the diagonals of a rhombus bisect each other at right angles.
Let ABCD be a rhombus whose diagonal AC and BD intersect at the point O.
As we know that the diagonals of a parallelogram bisect each other and rhombus is a parallelogram.
So, OA=OC and OB=OD.
From ∆ COB and ∆ COD we get,
CB = CD (sides of rhombus) and
CO is common in both the triangles.
So, OB = OD
Therefore, by SSS theorem.
∆ COB ≅ ∆ COD
COB = COD
COB + COD = 180° (Linear pair of angles)
Thus, COB = COD = 90°
Hence, the diagonals of a rhombus bisect each other at right angles.
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