In this chapter we will prove one of the universal law of triangles that states that ” the sum of two sides of triangle is always greater than twice the median “.

The below proof is important as it is directly asked in math exams. So make sure you understand the below steps.

**Given:**

Given above is the triangle ABC.

AM is the median that divide the side BC into two equal halves.

i.e. BM = MC.

**Construction:**

Extend median line MO such that AM = MO.

Also, join the point OC to form the triangle OMC.**To Prove:**

Prove that AB + AC > 2AM

**Solution**

Take** triangle MAB and MOC**

AM = MO { construction }

BM = MC { given }

∠AMB = ∠CMO { vertically opposite angles }

By** SAS congruency condition**, both triangles MAB and MOC are congruent.

i.e. \mathtt{\triangle MAB\ \cong \triangle MOC}

Since both triangles are congruent, we can say that;**AB = OC**

Now **take triangle ACO;**

We know that in triangle, sum of two sides is always greater than third side.

AC + CO > AO

We have already proved that CO = AB and AO = AM + MO

Putting the values;

AC + AB > AM + MO

AC + AB > AM + AM**AC + AB > 2 (AM)**

Hence, we proved that the sum of two sides is greater than twice the median.

**Next chapter** : **Isosceles triangle property**