Question

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

Answer 1

Given: A quadrilateral $ABCD$, in which $BM\perp AC$ and $DN\perp AC$ and $BM=DN$.
To prove: $AC$ bisects $BD$; or $DO=BO$
Proof:
Let $AC$ and $BD$ intersect at $O$.
Now, in $\Delta OND$ and $\Delta OMB$, we have:
$\angle OND=\angle OMB$ (${90}^{\circ }$ each)
$\angle DON=\angle BOM$
(Vertically opposite angles)
Also, $DN=BM$ (Given)
i.e., $\Delta OND\cong \Delta OMB$ (AAS congruence rule)
$OD=OB$ (From corresponding parts of congruent triangles)
Hence, $AC$ bisects $BD$.