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Question

In the adjoining figure, BM ⊥ AC and DN ⊥ AC. If BM = DN, prove that AC bisects BD.

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Solution

Given: A quadrilateral ABCD, in which BM ​⊥ AC and DN ⊥ AC and BM = DN.
To prove: AC bisects BD; or DO = BO

Proof:
Let AC and BD intersect at O.
Now, in ∆OND and ∆OMB, we have:
∠OND = ∠OMB (90o each)
∠DON = ∠ BOM (Vertically opposite angles)
Also, DN = BM (Given)
i.e.,
∆OND ≅ ∆OMB (AAS congurence rule)
∴ OD = OB (CPCT)
​Hence, AC bisects BD.

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