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

To do:

We have to show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Solution:

Let us draw a line $l$ and mark a point $P$ on it.

Now let us draw a perpendicular line $AB$ on $l$ and let us point a point $C$ on line $l$ and join $A$ to $C$

Let us now consider $\triangle ABC$,

We have,

$\angle B = 90^o$

We know that,

We know that,

The sum of the interior angles of a triangle is always equal to $180^o$

This implies

In $\triangle ABC,$

$\angle A+\angle B+\angle C = 180^o$

We have, 

$\angle B=90^o$

$\angle A+90^o+\angle C = 90^o$

$\angle A+\angle C=180^o-90^o$

$\angle A+\angle C=90^o$

Therefore, 

$\angle A$ and $\angle C must be acute angles

This implies,

$\angle A

We know that,

The side opposite to the larger angle is always larger

Therefore,

$AB

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Updated on: 10-Oct-2022

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