If ABCD is a quadrilateral whose diagonals AC and BD intersect at O, prove that
(i) `(AB + BC + CD +DA) gt (AC + BD),`
(ii) `(AB + BC + CD + DA) lt 2(AC + BD).`

Explanation

(i) We know that the sum of any two sides of a triangle is greater than the third.
In `triangle ABC, " we have " AB + BC gt AC.`
In `triangle ACD, " we have " CD + DA gt AC.`
In `triangle BCD, " we have " BC + CD gt BD.`
In `triangle ABD, " we have " DA + AB gt BD.`
adding these inequalities, we get
`2(AB + BC+CD+DA) gt 2(AC +BD).`
`therefore AB+BC+CD+DA gt AC +BD`
(ii) In `triangle AOB, " we have " OA + OB gt AB.`
In `triangle BOC, " we have " OB+OC gt BC. `
In `triangle COD, " we have " OC + OD gt CD.`
In `triangle DOA, " we have " OD + OA gt DA.`
Adding these inequalities, we get
`2(OA+OC)+2(OB+OD) gt AB + BC +CD +DA`
`rArr 2(AC +BD) gt AB + BC + CD +DA`
`rArr AB +BC +CD +DA lt 2(AC +BD).`