Question #aabf4

1 Answer
Jun 17, 2015

Use a proof by contradiction to show that no other line segment can be shorter than the perpendicular one.

Explanation:

Take PP as a point not on the line ABAB. Take QQ as the point on ABAB such that the line segment PQPQ is perpendicular to ABAB.

We're going to use proof by contradiction, meaning that we are going to assume the opposite of the statement we are trying to prove and prove that it is impossible.

Assume that there is a different line segment from PP to ABAB that is shorter than PQPQ - let's call it PRPR, where RR is a point on ABAB. This means that the length PR < PQPR<PQ. Since PRPR is different than PQPQ, Q != RQR.

This creates a right triangle PQRPQR, with a right angle at PQRPQR.

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Since we have a right triangle, we know that we can use the Pythagorean theorem, a^2 + b^2 = c^2a2+b2=c2 to find the length of each side. In this case, our hypotenuse is PRPR, so PQ^2+QR^2=PR^2PQ2+QR2=PR2.

Since QQ and RR are not the same point, QRQR is a non-zero, positive length. Likewise, PP is not on ABAB, so PQPQ and PRPR are non-zero positive lengths as well.

But, if each side is positive, PQ^2 = PR^2 - QR^2PQ2=PR2QR2, so PQ^2 < PR^2PQ2<PR2 and PQ < PRPQ<PR. This contradicts our original definition of PRPR!

Since we have proved that it is impossible for there to be a shorter line segment than the perpendicular line segment, we have proved that the perpendicular line segment is the shortest one.