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# 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

**☛ Related Questions:**

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