Show that in a quadrilateral ABCD `AB+BC+CD+DA gt AC + BD`

Explanation

Given ABCD is a quadrilateral

Construction
join diagonals AC and BD
To show ` AB+BC+CD+DA gt AC+BD`
In `Delta ABC , AB + BC gt AC" "...(i) `
[sum of two sides of a triangle is greater than the third side]
In `Delta BCD " "....(ii) ` [sum of two sides of a triangle is greater than the third side]
In `Delta CDA, " " CD+DA gt AC" "...(iii)`
[sum of two sides of a triangle is greater than the third side]
In `Delta DAB " " DA+ AB gt BD" "....(iv)`
[sum of two sides of a triangle is greater than the third side]
On adding Eqs,(i),(ii),(iii) and (iv), we get
`2(AB+BC+CD+DA) gt 2 (AC+BD)`
`rArr " " AB+BC+CD+DA gt AC +BD `