Draw `DE bot AB and CF bot AB` produced
In ` triangleAED and tirangleBFC`
` triangle AED cong triangleBFC`
AE = BF ( corresponding parts of congruent triangles) …(1)
Now L.H.S= `AC^(2) + BD^(2)`
`(AF^(2)+CF^(2))+ (DE^(2) + BE^(2))` ( by Pythoagoras theorem)
`AB^(2)+BF^(2)+2AB.BF+BC^(2)-BF^(2)+AD^(2)-AE^(2)+AB^(2)+AE^(2)-2AB.AE`
`=AB^(2)+2AB.AE+BC^(2)+AD^(2)+CD^(2)-2AB.AE( AE=BF and AB=CD)`
`AB^(2)+BC^(2)+CD^(2)+DA^(2)` Hence proved.