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 P as a point not on the line AB. Take Q as the point on AB such that the line segment PQ is perpendicular to AB.

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 P to AB that is shorter than PQ - let's call it PR, where R is a point on AB. This means that the length PR < PQ. Since PR is different than PQ, Q != R.

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

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

Since Q and R are not the same point, QR is a non-zero, positive length. Likewise, P is not on AB, so PQ and PR are non-zero positive lengths as well.

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

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.