show that the diogonals of a rhombus are perpendicular to each other
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⇒ Given :- ABCD is a rhombus
AC and BD are diagonals of rhombus intersecting at O.
⇒ To prove :- ∠BOC = ∠DOC = ∠AOD = ∠AOB = 90°
⇒ Proof :- All Rhombus are parallelogram, Since all of its sides are equal.
AB = BC = CD = DA ────(1)
The diagonal of a parallelogram bisect each other
Therefore, OB = OD and OA = OC ────(2)
In ∆ BOC and ∆ DOC
BO = OD [ From 2 ]
BC = DC [ From 1 ]
OC = OC [ Common side ]
∆ BOC ≅ ∆ DOC [ By SS congruency criteria ]
∠BOC = ∠DOC [ C.P.C.T ]
∠BOC + ∠DOC = 180° [ Linear pair ]
2∠BOC = 180° [ ∠BOC = ∠DOC ]
∠BOC = 180°/2
∠BOC = 90°
∠BOC = ∠DOC = 90°
Similarly, ∠AOB = ∠AOD = 90°
Hence, ∠BOC = ∠DOC = ∠AOD = ∠AOB = 90°
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