## Prove that in a triangle the sum of the medians is less than the perimeter. please answer with a diagram. ​

Question

Prove that in a triangle the sum of the medians is less than the perimeter. please answer with a diagram. ​

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3 weeks 2021-11-15T05:00:51+00:00 2 Answers 0 views 0

1. Figure is in the attachment

1)

Let AD,BE & CF be the three medians of a ∆ABC.

WE KNOW THAT THE SUM OF ANY TWO SIDES OF A TRIANGLE IS GREATER THAN TWICE THE MEDIAN DRAWN TO THE THIRD SIDE.

Hence, the perimeter of a triangle is greater than the sum of its three medians.

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2)

Use the result that the perpendicular drawn from a point( outside the line ) to a line is shorter( in length) than a line segment drawn from that point to the line and then add all three cases.

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Solution:

Consider ABC is a triangle and AL, BM and CN are the altitudes.

To Prove:

AL+BM+CN

Proof:

We know that the perpendicular AL drawn from the point A to the line BC is shorter than the line segment AB drawn from the point A to the line BC.

ALBMCN

AL+BM+CN

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3)

Given: A ∆ABC in which AD is median.

To Prove:

Construction:

Join EC.

Proof:

BD=CD. (D midpoint of BC)

∆ADB congruent ∆ EDC (by SAS)

AB=EC (by CPCT)

Now ,in ∆AEC, we have AC+EC>AE

[SINCE, SUM OF ANY TWO SIDES OF A TRIANGLE IS GREATER THAN THE THIRD SIDE]