Question
In the adjoining figure, $\text{BM}\perp\text{AC}$ and $\text{DN}\perp\text{AC}.$ If $BM = DN$, prove that $AC$ bisects $BD$.
✓
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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