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

^{0 }]

MX = MX [common]

so, triangle MPX is congruent to triangle MQX [SAS]

so, MP = MQ = [CPCT]

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