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

Given: l is a straight line and A is a point not lying on l. AB⊥ l and C is a point on l.
To prove: AB < AC
Proof: In ∆ABC,
∠B = 90° 

Since, C can lie anywhere on l (other than M)
So, AB is the shortest of all line segments drawn from A to l.

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 first we draw a line L & from point p draw two line segments PM & PN.


angle N = 90o

angle P+ angle M + angle N= 180o (angle sum property of triangle)

angle P + angle M=90that means angle M<90o (angle opp. to smaller side is smaller)

PN<PM(side opp. to smaller angle is smaller) Hence proved.

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May dis 1 help u....!!


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