Question

abcd is a quadrilateral prove ( ab+bc+cd+da) > ( ac+bd)

Answer 1

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

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

 AB + BC > AC  ------------------------- (i)

 BC + CD > BD ------------------------ (ii)

 AD + DC > AC ----------------------- (iii)

 AB + AD > BD ----------------------- (iv)

On combining (i),(ii),(iii),(iv) we get

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

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

= (AB + BC + CD + AD) > (AC + BD).

Hope this helps!

Answer 2

In triangle ACB AB + BC is greater than AC .....1
In triangle BCD BC+ DC is greater than DB. ...2
In triangle ACD AD+DC is greater than AC......3
In triangle DAB DA +AB is greater than DB....4
From equations 1 2 3 and 4
AB+BC+BC+DC+DC+AD+AD+AB > AC+DB+AC+DB
=2AB+2BC+2DC+2AD>2AC+2BC
=AB+BC+DC+AD>AC+BC......
Hence proved....