Prove that in a triangle the sum of the medians is less than the perimeter. please answer with a diagram.
Answers
Answer:
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.
AB+AC>2AD;. AB+BC>2BE & BC+AC>2CF.
Adding We get,
2(AB+BC+AC) >2(AD+BE+CF)
(AB+BC+AC) >(AD+BE+CF)
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
On adding equation 1,2,3
AL+BM+CN
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3)
Given: A ∆ABC in which AD is median.
To Prove:
AC+AB>2AD
Construction:
Produce AD to E, such that AD=DE.
Join EC.
Proof:
In ∆ADB & ∆ EDC
AD=ED (by construction)
angleADB=angleEDC. (vertically opposite angle)
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]
AC+EC>AD+DE. (AE=AD+DE)
AC+AB>2AD (AD=ED & EC=AB)
Thus, the sum of any two sides of a triangle is greater than twice the median with respect to the third side.
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Step-by-step explanation:
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