Question

ABCD is a quadrilateral in which diagonals AC and BD intersect at O .show that AB+BC+CD+DA>AC=BD

Answer 1

hiiii mate

ABCD is a quad.Its diagonals are AC and BD.

In triangle ACB, AB + BC > AC - 1

In triangle BDC, BC + CD > BD - 2

In triangle ACD, AD + DC > AC -3

In triangle BAD, AB + AD > BD -4

Adding 1,2,3 and 4,

AB + BC + BC + CD + AD + DC + AB + AD > AC + BD + AC + BD

2AB + 2BC + 2CD + 2AD > 2AC + 2BD

AB + BC + CD + AD > AC + BD.

Answer 2

In ∆ABC,

AB+BC>AC ,.........eq(1),
(since in any ∆ the sum of any two sides is always greater than the third side),

similarly
in ∆ BCD

BC+CD>BD,...............eq(2),


in ∆CDA,

CD+DA>AC,...........eq(3),


similarly
in ∆DAB,

DA+AB>AD,..................eq(4),



now adding all these equations , we have

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

then

AB+BC+CD+DA>AC+BD