Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side.
Solution:
Consider a triangle ABC
AD is a median
We have to prove that the sum of any two sides of a triangle is greater than twice the median with respect to the third side.
Extend AD upto E such that AD = DE and join CE.
Considering triangles ABD and ECD,
Given, AD = DE
Since AD is the median, BD = CD
We know that the vertically opposite angles are equal.
∠ADB = ∠CDE
By SAS criterion, the triangles ABD and ECD are congruent.
By CPCTC,
AB = EC
In triangle AEC,
We know that in a triangle, the sum of two sides is greater than the third side.
AC + EC > AE
From (1),
AC + AB > AE
From the figure,
AE = AD + DE
Also, AD = DE
So, AE = AD + AD
AE = 2 AD
So, AC + AB > 2AD
Therefore, the sum of any two sides of a triangle is greater than twice the median with respect to the third side.
✦ Try This: In △ABC,AB=AC and AD is the perpendicular bisector of BC. Show that △ADB≅△ADC
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 7
NCERT Exemplar Class 9 Maths Exercise 7.4 Problem 10
Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side
Summary:
It is proven that 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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