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)`

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 `EOF || AB` and `LOM||AD`.
Proof: `EOF||AB` and `DA` cuts them.
`:./_DEO=/_EAB=90^@`(corresponding /_).
`:.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(rec. ABCD)`.

Again, `ar(/_\AOB)+ar(/_\COD)`
`=(1/2xxABxxOL)+(1/2xxDCxxOM)`
`=1/2xxABxx(OL+OM)` [`:.DC=AB`]
`=(1/2xxABxxOL)+(1/2×DC×OM)`
`=1/2xxABxx(OL+OM)`
`=1/2xxar(rect. ABCD)`.
`:.ar(/_\AOD)+ar(/_\BOC)=ar(/_\AOB)+ar(/_\COD)`.