Draw `AL bot BC and MC bot BC`
In `triangleOLA and triangleOMD`
` angleALO = angle= 90^(@)`
and ` angle AOL = angleDOM` ( vertically opposite angle)
`angleOLA ~ angleOMD` ( AAA similarity criterion)
`(AL)/(DM) = (AO)/(DO)` …(1)
Now, ` (ar(triangleABC))/(ar(triangleDBC))= (1/2xx(BC)xx(AL))/(1/2xx(BC)xx(DM))(AL)/(DM)= (AO)/(DO)`
Hence , ` (ar(triangleABC)/(ar(triangleDBC))= (AO)/(DO)` Hence proved.