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Question
In the adjoining figure, BM ⊥ AC and D DN ⊥ AC. If BM = DN, prove that AC bisects BD.
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Solution
ANSWER:
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
(90 o each)
∠DON = ∠ BOM
(Vertically opposite angles)
Also, DN = BM
(Given)
i.e., ∆OND ≅ ∆OMB
(AAS congurence rule)
∴ OD = OB
(CPCT)
Hence, AC bisects BD.
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
(90 o 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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