Let `AD = 3x cm and DB=2x cm`. Then,
`AB= (AD+DB)=(3x+2x) cm =5cm`.
In `Delta ABC and Delta DBE` we have
`angle CAB= angle EDB` [ corresponding `angle`]
and ` angle AC~ angle DEB` [ corresponding `angle`]
` :. Delta ABC~ Delta DBE`[ by AA- similarity]
`rArr (ar (Delta ABC))/(ar (Delta DBE))=((AB)^(2))/((DB)^(2))=((5x))/((2x)^(2))=(25x^(2))/(4x^(2)) =(25)/(4)`
`rArr ar (Delta ABC) : ar (Delta DBE)= 25.4`