# In any triangle ABC, BP and CQ are perpendiculars on any line through A and M is the mid-point of the side BC. Show that MP=MQ

In the given figure,
M is the midpoint of BC, that means = BM = MC.
Also, BP $\perp$ AQ , CQ $\perp$ AQ
​From point M, draw MX $\perp$ AQ
​Since BP, MX and CQ are perpendicular to same line AQ, so they are parallel to each other.
Since M is the midpoint of BC, so intercepts made by these parallel lines on PQ will also be equal.
Hence, PX = XQ.
In triangle MPX and MQX,
PX = XQ    [proved above]
$\angle$PXM = $\angle$QXM   [each 90]
MX = MX   [common]
so, triangle MPX is congruent to triangle MQX [SAS]
so, MP = MQ = [CPCT]

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