A straight line parallel to side BC of triangle ∆ABC intersects sides AB and AC at points D and E respectively. If AD : BD = 3 : 5, then what is the ratio of the area of triangle ∆ADE to the area of trapezium DBCE?

Q.A straight line parallel to side BC of triangle ∆ABC intersects sides AB and AC at points D and E respectively. If AD : BD = 3 : 5, then what is the ratio of the area of triangle ∆ADE to the area of trapezium DBCE?

A perpendicular AP is drawn from BC which intersects DE at point Q. Given, \( \frac{AD}{BD} = \frac{3}{5} \) ∴ \( \frac{AQ}{PQ} = \frac{AD}{BD} = \frac{3}{5} \) i.e., \( PQ = \frac{5}{3} AQ \) Again, \( \frac{BD}{AD} + 1 = \frac{5}{3} + 1 \) i.e., \( \frac{BD + AD}{AD} = \frac{5 + 3}{3} \) i.e., \( \frac{AB}{AD} = \frac{8}{3} \) ∴ \( \frac{AP}{AQ} = \frac{BC}{DE} = \frac{8}{3} \) ∴ \( BC = \frac{8}{3} DE \) ∴ Area of triangle ∆ADE : Area of trapezium DBCE = \( \frac{1}{2} \times DE \times AQ : \frac{1}{2} \times (DE + BC) \times PQ \) = \( \frac{1}{2} \times DE \times AQ : \frac{1}{2} \times (DE + \frac{8}{3} DE) \times \frac{5}{3} AQ \) = \( 1 : (1 + \frac{8}{3}) \times \frac{5}{3} \) = \( 1 : \frac{11}{3} \times \frac{5}{3} \) = \( 9 : 55 \)