Question

Prove that, of all line segments that can be drawn to a given line, from a point, not lying on it, the perpendicular line segment is the shortest.

Answer 1

Given: l is a straight line and A is a point not lying on l.
AB is perpendicular to line l.
C is a point on l.
To prove: $AB<AC$
Proof:
$$\begin{gathered} \angle B=90° \\ In△ABC,\mathrm{we}\mathrm{have}: \\ \angle C<90°\left[\angle A+\angle B+\angle C=180 \\ \Rightarrow \angle A+\angle C=90° \\ \Rightarrow \angle C<90°\right] \\ \Rightarrow \angle C<\angle B \\ \Rightarrow AB<AC(\mathrm{Greater}\mathrm{angle}\mathrm{has}\mathrm{a}\mathrm{greater}\mathrm{side}\mathrm{opposite}\mathrm{to}\mathrm{it}) \end{gathered}$$
Since, C can lie anywhere on l, AB is the shortest of all line segments drawn from A to l.