Question
In the figure, $BM$ and $DN$ are both perpendiculars on $AC$ and $BM = DN.$ Prove that $AC$ bisects $BD.$
✓
Answer
In $\triangle BMR$ and $DNR$
$BM = DN$
$\angle BMR = \angle DNR = 90^\circ $
$\angle BRM = \angle DRN = ...($vertically opposite angles$)$
Hence, $\angle MBR = \angle NDR ...($sum of angles of a triangle $= 180^\circ )$
$\triangle BMR ≅ \triangle DNR ...(\text{ASR}$ criteria$)$
Therefore, $BR = DR$
So, $AC$ bisects $BD.$
$BM = DN$
$\angle BMR = \angle DNR = 90^\circ $
$\angle BRM = \angle DRN = ...($vertically opposite angles$)$
Hence, $\angle MBR = \angle NDR ...($sum of angles of a triangle $= 180^\circ )$
$\triangle BMR ≅ \triangle DNR ...(\text{ASR}$ criteria$)$
Therefore, $BR = DR$
So, $AC$ bisects $BD.$
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