Question

if ABCD is a quadrilateral whose diagonals AC and BD intersect at O, prove that (1)(AB+BC+CD+DA)>(AC+BD)

Answer 1

In triangle ABC by the triangle inequality we have AB + BC > AC.
In triangle ADC by the triangle inequality we have AD + CD > AC.
In triangle BCD by the triangle inequality we have BC + CD > BD.
In triangle BAD by the triangle inequality we have AB + AD > BD.

Add these up

(AB + BC) + (AD + CD) + (BC + CD) + (AB + AD) > AC + AC + BD + BD
2AB + 2BC + 2AD + 2CD > 2AC + 2BD
2(AB + BC + AD + CD) > 2(AC + BD)
2(perimeter) > 2(sum of diagonals)

Answer 2

ABCD is a quadrilateral and AC, and BD are the diagonals.

Sum of the two sides of a triangle is greater than the third side.

So, considering the triangle ABC, BCD, CAD and BAD, we get

AB + BC > AC

CD + AD > AC

AB + AD > BD

BC + CD > BD

Adding all the above equations,

2(AB + BC + CA + AD) > 2(AC + BD)

⇒ 2(AB + BC + CA + AD) > 2(AC + BD)

⇒ (AB + BC + CA + AD) > (AC + BD)

⇒ (AC + BD) < (AB + BC + CA + AD)