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Answer
Given: A quadrilateral ABCD, in which $\text{BM}\perp\text{AC}$ and $\text{DN}\perp\text{AC}$ and BM = DN.
To prove: AC bisects BD; or DO = BO
Proof:
Let AC and BD intersect at O.
Now, in $\triangle\text{OND}$ and $\triangle\text{OMB},$ we have:
$\angle\text{OND}=\angle\text{OMB}$ (90° each)
$\angle\text{DON}=\angle\text{BOM}$ (Vertically opposite angles)
Also, $\text{DN = BM}$ (Given)
i.e., $\triangle\text{OND}\cong\triangle\text{OMB}$ (AAS congrurence rule)
$\therefore\text{OD = OB}$ (C.P.C.T.)
Hence, AC bisects BD.
To prove: AC bisects BD; or DO = BO
Proof:
Let AC and BD intersect at O.
Now, in $\triangle\text{OND}$ and $\triangle\text{OMB},$ we have:
$\angle\text{OND}=\angle\text{OMB}$ (90° each)
$\angle\text{DON}=\angle\text{BOM}$ (Vertically opposite angles)
Also, $\text{DN = BM}$ (Given)
i.e., $\triangle\text{OND}\cong\triangle\text{OMB}$ (AAS congrurence rule)
$\therefore\text{OD = OB}$ (C.P.C.T.)
Hence, AC bisects BD.
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