VerifiedHint: Draw a figure and use the properties of triangles to get the necessary result.
Complete step-by-step answer:
Consider a line $l$ on which two points Y and Z lie,
Now, suppose a perpendicular is drawn from a point X on the line $l$ at Y and another line segment from X meets the line $l$ at some arbitrary point Z (such that Z is not the same as Y).
To prove: XY<XZ
Proof:
Consider the $\vartriangle XYZ$
$\angle XYZ = {90^\circ} \\$
Also in $\vartriangle XYZ$
$\angle YXZ + \angle XZY + \angle XYZ = {180^\circ}$..........($\because$ Angle Sum Property)
Putting $\angle XYZ = {90^\circ} \to (1)$
We get,
$\angle YXZ + \angle XZY = {90^\circ} \to (2)$
Now, clearly $\angle XZY < {90^\circ} \to (3)$ (for the triangle to exist)
So, from equations (1) and (3), we can say that,
$\angle XZY < \angle XYZ$
So, we can say XY < XZ (side opposite to greater angle is longer)
Hence Proved.
Hence we can say that XY is the shortest compared to all line segments from point X to line $l$.
Note: Such problems can be solved by using properties of triangles like the side opposite to greater angle is the longer. Remember to draw a figure for solving more easily.
