Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest
Answers
AShow that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Solution:
We know that in a triangle if one angle is 90 degrees, then the other angles have to be acute.
Show 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, that is, 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 ΔPNM, ∠N = 90o
∠P + ∠N + ∠M = 180o (Angle sum property of a triangle)
∠P + ∠M = 90o
Clearly, ∠M is an acute angle.
∠M < ∠N
PN < PM (The side opposite to the smaller angle is smaller)
Similarly, by drawing different line segments from P to l, it can be proved that PN is smaller in comparison to all of 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.nswer:
Step-by-step explanation: