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See diagram

We have a straight line  AB.
Take a point P not on AB, but some distance away.  We draw PC the perpendicular on to AB meeting AB at C.  We draw another line from P to D, where D lies on AB.  We show that PD > PC, for any point D other than C.

In the triangle PCD, use the Pythagorean law for sides:
PC² + CD² = DP²

Since CD² is positive and adds to PC²,    PC²+CD² > PC²
Hence DP² > PC²

So  DP > PC

SO PC, the perpendicular distance from an external point to a line segment, is the shortest.

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