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

Hi!
 
 
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.
 
Cheers!

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

here,

angle N = 90o

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

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

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

this question is also on meritnation u will take help frm there.

May dis 1 help u....!!


 

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