Question

Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Answer 1

Consider a line segment $\text{'}\mathrm{l}\text{'}$ on which$‘\mathrm{B}’$and$‘\mathrm{C}’$are the two points.

Now, consider a point $‘\mathrm{A}’$ away from the given line such that $\mathrm{AB}⟂\mathrm{BC}$.

In $∆\mathrm{ABC}$, by angle sum property, we have

$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}={180}^0$

$\angle \mathrm{A}+{90}^0+\angle \mathrm{C}={180}^0$ (given $\angle \mathrm{B}={90}^0$)

⇒$\angle \mathrm{A}+\angle \mathrm{C}={180}^0-{90}^0$

or$\angle \mathrm{A}+\angle \mathrm{C}={90}^0$

It is evident that $\angle \mathrm{A}$ $<{90}^0$ and $\angle \mathrm{C}$ $<{90}^0$ are the acute angles.

We know that, side opposite to the greater angle is the longer side. $\angle \mathrm{B}$ is the greatest angle in $∆\mathrm{ABC}$ so the hypotenuse $\mathrm{AC}$ is the longest side. So, $\mathrm{AB}$ is the shortest distance in this case.

Thus, all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Hence Proved