In adjoining figure, AP ⊥ BC, AD || BC, then find A( ABC):A( BCD).

Given , AP ⊥ BC, and AD || BC. ABC and BCD has same base BC.
Areas of triangles with equal bases are proportional to their corresponding heights.
Since AP is the perpendicular distance between parallel lines AD and BC, height of ABC and height of BCD are same.
A( ABC) /A( BCD) = AP/AP = 1
Hence A( ABC): A( BCD) = 1:1