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Show that in a quadrilateral ABCD `AB+BC+CD+DA lt 2 (BD + AC)`

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Given : A quadrilateral ABCD.
To prove : `AB+BC+CD+DA lt 2(BD+AC)`
Proof : In `DeltaAOB`, we have
`implies" "OA+OB gt AB` ...(1)
( `:'` sum of any two sides of a triangle is greater then the third side) In `DeltaBOC`, we have
`OB+OC gt BC` (same reason) ...(2)
In `DeltaCOD`, we ahve
`OC+OD gt CD` (same reason) ...(3)
In `DeltaDOA`, we have
`OD+OA gt DA` (same reason) ...(4)
Adding (1), (2), (3) and (4), we get
`OA+OB+OB+OC+OC+OD+OD+OA gt AB + BC+CD+DA`
`implies" "2(OA+OB+OC+OD) gt AB+BC+CD+DA`
`implies" "2{(OA+OC)+(OC+OD)} gt AB+BC+CD+DA`
`implies" "2(AC+BD) gt AB+BC+CD+DA`
`implies" "AB+BC+CD+DA lt 2 (BD+AC)`
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