Question

prove that the sum of two sides of a triangle is greater than the third side in a triangle

Answer 1

it is very easy

let in triangle ABC

AC is longer side AND AB is altitude and BC is base

so, angle A = BC [ EQUAL TO OPPOSITE SIDE]

ANGLE B = AC

AND ANGLE C = AB

SO ANGLE B IS GREATER BECAUSE HIS SIDE AC IS GREATER

Answer 2

Given ---
A ∆ABC

To prove ---
AB + AC > BC

Construction ---
Produce BA to D such that AD = AC. Join CD

Proof ---

In ∆ACD, we have

AD = AC ....(by construction)...(Equation 1)

=> ∠ACD = ∠ADC ....(∠s opposite to equal sides are equal)

=> ∠BCD > ∠ACD ....(whole is greater than a part)

=> ∠BCD > ∠ADC ....(because ∠ACD = ∠ADC)

Now, since ∠ADC = ∠BDC

=> ∠BCD > ∠BDC

=> BD > BC ....(sides opposite to larger ∠ is longer)

=> BA + AD > BC

=> BA + AC > BC

=> AB + AC > BC
$proved$

Similarly, we can prove that,
AB + BC > AC
and,
BC + AC > AB

Hence, it's proved that the sum of any two sides of a ∆ is greater than the third side.

Hope this answer helps you. :-)