Show that of all lines segment …

Let the given point be P and line be l
We draw two line segments PN & PM 
such that 
PM{tex} \perp{/tex} MN
We have to prove: PM > PN
In {tex}\Delta \mathrm{P} \mathrm{NM}{/tex}
{tex}\angle \mathrm{P}+\angle \mathrm{N}+\angle \mathrm{M}=180^{\circ}{/tex} (Angle sum property of triangle)
{tex}\angle \mathrm{P}+90^{\circ}+\angle \mathrm{M}=180^{\circ}{/tex} (As PN {tex} \perp{/tex}MN, {tex}\angle N=90^{\circ}{/tex}
{tex}\angle \mathrm{P}+\angle \mathrm{M}=180^{\circ}-90^{\circ}{/tex}
{tex}\angle \mathrm{P}+\angle \mathrm{M}=90^{\circ}{/tex}

Since angle can't be 0 or negative
Hence,
{tex}\angle \mathrm{M}<90^{\circ}{/tex}
{tex}\angle \mathrm{M}<\angle \mathrm{N}{/tex}
PN < PM (Side opposite to the greater angle is longer)
{tex}\therefore{/tex} Perpendicular line segment is the shortest.
Hence proved.