Question

Question #aabf4

Answer 1

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`.

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.