Question

Prove that in a quadrilateral the sum of all the sides is greater than the sum of its diagonals.

Answer 1

Given : In quadrilateral ABCD, AC and BD are its diagonals,

Tor prove : $AB+BC+CD+DA>AC+BD$

Proof : In $\Delta ABC,$

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

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

Similarly, in $\Delta ADC,$

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

$In\Delta ABD,$

$AB+DA>BD...\left(iii\right)$

$In\Delta BCD,$

$BC+CD>BD...\left(iv\right)$

$\text{Adding(i), (ii), (iii) and (iv)}$

$2(AB+BC+CD+DA)>2AC+2BD$

$\Rightarrow (AB+BC+CD+DA)>2(AC+BD)$

$\therefore AB+BC+CD+DA>AC+BD$