Revision: Democratic Politics 2 >> Gender, Religion and Caste Social Science English Medium Class 10 CBSE

Given: ΔABC is on which AD, BE and CF are its medians.

To Prove: We know that the sum of any two sides of a triangle is greater than twice the median bisecting the third side.

Therefore, AD is the median bisecting BC

⇒ AB + AC > 2AD    ...(i)

BE is the median bisecting AC   ...(ii)

And CF is the median bisecting AB

⇒ BC + AC > 2EF    ...(iii)

Adding (i), (ii) and (iii), we get

(AB + AC) + (AB + BC) + (BC + AC) > 2. AD + 2 . BE + 2 . BE + 2 . CF

⇒ 2 (AB + BC + AC) > 2 (AD + BE + CF)

⇒ AB + BC + AC > AD + BE + CF.

Theorem : If two sides of a triangle are unequal, the angle opposite to the longer side is larger (or greater).
Activity :  Draw a line-segment AB. With A as centre and some radius, draw an arc and mark different points say P, Q, R, S, T  on it.  Join each of these points with A as well as with B in following fig . 

Observe that as we move from P to T, ∠ A is becoming larger and larger. The length of the side is also increasing; that is ∠ TAB > ∠ SAB > ∠ RAB > ∠ QAB > ∠ PAB and TB > SB > RB > QB > PB.
Now, draw any triangle with all angles unequal to each other. Measure the lengths of the sides in following fig. 

Observe that the side opposite to the largest angle is the longest. In Fig , ∠ B is the largest angle and AC is the longest side.

Theorem : In any triangle, the side opposite to the larger (greater)  angle is longer. 
Now take a triangle ABC and in it, find AB + BC, BC + AC and AC + AB.  
You will observe that AB + BC > AC, BC + AC > AB  and AC + AB > BC.

Theorem : The sum of any two sides of a triangle is greater than the third side. 

In above fig.  the side BA of ∆ ABC has been produced to a point D such that AD = AC.