Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Solution:
Let us take a line l and from point P (i.e., not on line l), draw two line segments PN and PM. Let PN be perpendicular to line l and PM is drawn at some other angle.
In $\triangle P N M$,
$\angle \mathrm{N}=90^{\circ}$
$\angle \mathrm{P}+\angle \mathrm{N}+\angle \mathrm{M}=180^{\circ}$ (Angle sum property of a triangle)
$\angle P+\angle M=90^{\circ}$
Clearly, $\angle M$ is an acute angle.
$\therefore \angle \mathrm{M}<\angle \mathrm{N}$
$\Rightarrow$ PN < PM (Side opposite to the smaller angle is smaller)
Similarly, by drawing different line segments from $\mathrm{P}$ to $\mathrm{l}$, it can be proved that $\mathrm{PN}$ is smaller in comparison to them.
Therefore, it can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Let us take a line l and from point P (i.e., not on line l), draw two line segments PN and PM. Let PN be perpendicular to line l and PM is drawn at some other angle.
In $\triangle P N M$,
$\angle \mathrm{N}=90^{\circ}$
$\angle \mathrm{P}+\angle \mathrm{N}+\angle \mathrm{M}=180^{\circ}$ (Angle sum property of a triangle)
$\angle P+\angle M=90^{\circ}$
Clearly, $\angle M$ is an acute angle.
$\therefore \angle \mathrm{M}<\angle \mathrm{N}$
$\Rightarrow$ PN < PM (Side opposite to the smaller angle is smaller)
Similarly, by drawing different line segments from $\mathrm{P}$ to $\mathrm{l}$, it can be proved that $\mathrm{PN}$ is smaller in comparison to them.
Therefore, it can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.