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(N/A) Let us consider a point $P$ not on line $l$. Let $PM$ be the perpendicular from $P$ to line $l$,and let $N$ be any other point on line $l$ such

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(N/A) Let us consider a point $P$ not on line $l$. Let $PM$ be the perpendicular from $P$ to line $l$,and let $N$ be any other point on line $l$ such that $N 
eq M$.In $\Delta PMN$,since $\angle M = 90^o$,the sum of angles is $\angle M + \angle N + \angle P = 180^o$.This implies $\angle N + \angle P = 90^o$,so $\angle N Since $\angle N Therefore,$PM Since $N$ is an arbitrary point on line $l$ (other than $M$),this shows that $PM$ is shorter than any other line segment drawn from $P$ to line $l$.Thus,the perpendicular segment is the shortest line segment drawn from a point to a line.

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

Similar Questions

In the figure,$PR > PQ$ and $PS$ bisects $\angle QPR$. Prove that $\angle PSR > \angle PSQ$.

$ABC$ and $DBC$ are two isosceles triangles on the same base $BC$ (see figure). Show that $\angle ABD = \angle ACD$.

$AD$ and $BC$ are equal perpendiculars to a line segment $AB$ (see figure). Show that $CD$ bisects $AB$.

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$AB$ is a line segment and line $l$ is its perpendicular bisector. If a point $P$ lies on $l$,show that $P$ is equidistant from $A$ and $B$.

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