Question

In the figure, prove that:

$\left(i\right)CD+DA+AB+BC>2AC$

$\left(ii\right)CD+DA+AB>BC$

Answer 1

Given : In the figure, ABCD is a quadrilateral and AC is joined

To prove :

$\left(i\right)CD+DA+AB+BC>2AC$

$\left(ii\right)CD+DA+AB>BC$

Proof :

(i) In $\Delta ABC,$

$AB+BC>AC...\left(i\right)$

(Sum of two sides of a triangle is greater than its third side)

Similarly in $\Delta ADC,$

$CD+DA>AC...\left(ii\right)$

Adding (i) and (ii)

$CD+DA+AB+BC>AC+AC$

$\Rightarrow CD+DA+AB+BC>2AC$

(ii) In $\Delta ACD.$

$CD+DA+>CA$

(Sum of two sides of a triangle is greater than its third side)

Adding AB to both sides,

$CD+DA+AB>CA+AB$

But $CA+AB>BC\left(in\Delta ABC\right)$

$\therefore CD+DA+AD>BC$