Question

In quad ABCD AC and BD are diagonal PROVE 2(AC+BD)>AB+BC+CD+AD

Answer 1

● Quadrilaterals ●

In quadrilateral ABCD, AC and BD are diagonals.

To prove : - 2(AC + BD) > AB + BC + CD + AD

Now,

Let the intersection point of the diagonals be O.

We know that, in a triangle the sum of two sides is greater than the third side.

Then,

In ∆AOB, AO + BO > AB ......(i)
In ∆BOC, BO + CO > BC .....(ii)
In ∆COD, CO + DO > DC .....(iii)
In ∆AOD, AO + DO > AD .....(iv)

Adding (i), (ii), (iii) and (iv), we get ,

AO + BO + BO + CO + DO + AO + DO > AB + BC + CD + DA

= 2(AO + CO) + 2(BO + DO) > AB + BC + CD + DA

= 2AC + 2BD > AB + BC + CD + DA

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

Hence, Proved.