In the adjoining figure, ABCD is a quadrilateral. Its diagonals AC and BD intersect at point 'O'. Prove that :
(a) `AB+BC+CD+DA lt 2(AC+BD)`
(b) `AB+BC+CD+DA gt (AC+BD)`

Explanation

(a) We know that the sum of any two sides of a triangle is greater than the third side.
`{:( :'" In "DeltaOAB",",OA+OB gt AB,...(1)),("In "DeltaOBC",",OB+OC gt BC,...(2)),("In "DeltaOCD",",OC+OD gt CD,...(3)),("In "DeltaODA",",OD+OA gt DA,...(4)):}`
`OA+OB+OB+OC+OC+OD+OD+OA gt AB+BC+CD+DA`
`implies" "2OA+2OB+2OC+2OD gt AB+BC+CD+DA`
`implies" "2(OA+OC)+2(OB+OD) gt AB+BC+CD+DA`
`implies" "2AC+2BD gt AB+BC+CD+DA`
`implies" "2(AC+BD) gt AB+BC+CD+DA`
`implies" "AB+BC+CD+DA lt 2(AC+BD)`

(b) `{:("In "DeltaABC",",AB+BC gt AC,...(5)),("In "DeltaBCD",",BC+CD gt BD,...(6)),("In "DeltaCDA",",CD+DA gt AC,...(7)),("In "DeltaDAB",",DA+AB gt BD,...(8)):}`
Adding (5), (6), (7) and (8), we get
`AB+BC+BC+CD+CD+DA+DA+AB gt AC+BD+AC+BD`
`implies" "2(AB+BC+CD+DA) gt 2(AC+BD)`
`implies" "AB+BC+CD+DA gt AC+BD`