Question

Prove that the difference of any two sides of a triangle is less than the third side.

Answer 1

Given: A $\Delta ABC$
To prove:
(i) $AC–AB<BC$
(ii) $BC–AC<AB$
(iii) $BC–AB<AC$
Construction: Take a point $D$ on $AC$ such that $AB=AD$. Join $BD$.
Proof: In $\Delta ABD,AB=AD,\angle 1=\angle 2$ …(i)
In $\Delta ABD$, side $AD$ is produced to $C\angle 3>\angle 1$ …(ii)
(ext. angle is greater than one of the opposite interior angle)
Also in $\Delta BCD$
$\angle 2>\angle 4$ …(iii) (reason as above)
From (i) and (ii), we get
$\angle 3>\angle 2$ …(iv)
From (iii) and (iv), we get
$\angle 3>\angle 4$
$BC>CD$ (side opposite to greater angle is larger)
$CD<BC$
$AC–AD<BC$
$AC–AB<BC$ $[\because AD=AB]$
Similarly we can prove
$BC–AC<AB$ and $BC–AB<AC$
Hence proved.