Question

Prove that the sum of any two sides of a triangle is always greater than the third side.

Answer 1

Hint: Use the fact that in a triangle, the larger angle has a larger opposite side. Extend AC to point D such that CD = BC. Join BD. Observe in triangle ADB $\angle ABD>\angle ADB$. Apply the above-mentioned theorem and hence prove the result.

Complete step-by-step answer:

Given: A triangle ABC.
To prove: $AC+BC>AB$
Construction: Extend AC to a point D such that CD = CB. Join BD.
Proof:
In triangle BCD, we have
BC=DC
Hence $\angle CDB=\angle CBD$ (because equal sides of a triangle have equal opposite angles)
Now since $\angle ABD>\angle CBD$, we have
$\angle ABD>\angle CDB$.
Now in triangle ABD, we have
$\angle ABD>\angle CDB$
Hence we have $AD>AB$( because the side opposite to a larger angle is longer).
Now we have
AD = AC+CD
Since CD = BC, we have
AD = AC+BC.
Hence $AD>AB\Rightarrow AC+BC>AB$
Hence the sum of two sides of a triangle is larger than the third side.
Note: [1] The above inequality is strict, i.e. We cannot have a triangle in which the sum of sides is even equal to the third side.
[2] The above theorem is necessary for the existence of a triangle and is also sufficient for the existence of the triangle, i.e. if the sum of any two sides is greater than the third side, then the triangle with the given sides exists and if there exists a triangle then the sum of its any two sides is always greater than the third side.