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` and `O` is a point inside it. `O A ,O B,O C` and `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)` CONSTRUCTION : Draw `E O F A B` and `L O M A Ddot`

Explanation

Given: A rectangle `ABCD` and `O` is a point inside it. `OA`, `OB`, `OC` and `OD` have been joined.
To Prove: `ar(/_\AOD)+ar(/_\BOC)=ar(/_\AOB)+ar(/_\COD)`
Construction: Draw `EF∥AB` and `LOM∥AD`
Proof: `EF∥AB` and `DA` cuts them.
`:./_DEO=/_EAB=90^@` (corresponding `/_s`)
`∴OE_|_AD`
`:.OE_|_AD`
Similarly, `OF_|_BC`, `OL_|_AB` and `OM_|_DC`
`:.ar(/_\AOB)+ar(/_\BOC)`
`=(1/2xxADxxOE)+(1/2xxBCxxOF)=1/2ADxx(OE+OF)` [`:.BC=AD`]
`=1/2xxADxxEF=1/2xxADxxAB` [`:.EF=AB`]
`=1/2xxar(`rectangle `ABCD)`
Again, `ar(/_\AOB)+ar(/_\COD)`
`=(1/2xxABxxOL)+(1/2xxDCxxOM)=1/2xxABxx(OL+OM)` [`:.DC=AB`]
`=1/2xxABxx(OL+OM)=1/2xxar(`rectangle `ABCD)`
`∴ ar(/_\AOD)+ar(/_\BOC)=ar(/_\AOB)+ar(/_\COD)`
Hence Proved.