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A point O inside a rectangle A B C D ...

A point `O` inside a rectangle `A B C D` is joined to the vertices. Prove that the sum of the areas of a pair of opposite triangles so formed is equal to the sum of the other pair of triangles. Given: A rectangle `A B C D\ a n d\ O` is a point inside it. `O A ,\ O B ,\ O C\ a n d\ O D` have been joined. To Prove: `a r\ (A O D)+\ a r\ ( B O C)=\ a r\ ( A O B)+\ a r( C O D)`

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GIVEN A rect. ABCD and O is a point inside it. OA, OB, OC and OD have been joined.
TO PROVE `ar(triangleAOD)+ar(triangleBOC)=ar(triangleAOB)+ar(triangleCOD).`
CONSTRUCTION Draw EOF || AB and LOM||AD.
PROOF EOF||AB and DA cuts them.
`therefore angleDEO=angleEAB=90^(@)" "("corres". angle).`
`therefore OEbotAD.`

Similarly, `OF bot BC,OLbotAB and OM bot DC.`
`therefore ar(triangleAOB)+ar(triangleBOC)`
`=((1)/(2)xxADxxOE)+((1)/(2)xxBCxxOF)`
`=(1)/(2)ADxx(OE+OF) " "[therefore BC=AD]`
`=(1)/(2)xxADxxEF =(1)/(2)xxADxxAB" "[therefore EF=AB]`
`=(1)/(2)xxar("rec. ABCD")`.
Again, `ar(triangleAOB)+ar(triangleCOD)`
`=((1)/(2)xxABxxOL)+((1)/(2)xxDCxxOM)=(1)/(2)xxABxx(OL+OM)`
`[therefore DC=AB]`
`=((1)/(2)xxABxxOL)+((1)/(2)xxDCxxOM)=(1)/(2)xxABxx(OL+OM)`
`=(1)/(2)xxar("rect. ABCD")`.
`therefore ar(triangleAOD)+ar(triangleBOC)=ar(triangleAOB)+ar(triangleCOD)`.
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