Theorem 4

Prove that: From a point, outside a line, the perpendicular is the shortest from the point to the line.

Given: From a point $C$, $\overline{CD}$ is drawn perpendicular to AB meeting it in D and $\overline{CE}$ is any other segment meeting AB in E.

To prove: $m\overline{CD}$<$m\overline{CE}$

Explanation

• Given a point $C$ and a line $AB$, we draw a perpendicular $CD$ from point $C$ to line $AB$, and we draw another segment $CE$ from point $C$ to line $AB$ at point $E$. Our goal is to prove that $CD$ is the shortest segment from point $C$ to line $AB$.

• To do this, we consider an exterior angle $1$ of triangle $CDE$, which is the angle formed by extending one of the sides of the triangle (in this case, side $DE$) outside of the triangle. By the definition of exterior angles, angle $1$ is greater than the opposite interior angle $3$.

• We also know that angle $2$, formed by the perpendicular $CD$ and the line $AB$, is a right angle, because $CD$ is perpendicular to $AB$. Since angles $1$ and $2$ are linear pairs (add up to $180\degree$), we can conclude that angle $2$ is also greater than angle $3$.

• Therefore, we have shown that angle $1$ is greater than angle $3$, and angle $2$ is greater than angle $3$, which means that angle $2$ is the greatest of the three angles. Since the side opposite the greatest angle in a triangle is the longest side, we can conclude that $CE$ is longer than $CD$.

• Thus, we have proved that $CD$ is the shortest segment from point $C$ to line $AB$.