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