∫AE:DE = 2: 1

∴AE:AD=2:3

DE:AD= 1:3

Advertising split ∠BAC

∴∠BAE=∠CAD

∠∠ABE =∠C C。

∴△ABE∽△ACD(AA)

∴BE:CD=AE:AD=2:3

∠AEB=∠ADC

∴∠BED=∠BDE (complementary angles of equal angles are equal)

∴BE=BD

∴BD:CD=2:3

So BD: BC = 2: 5.

∫△BDE and△△△ Abd are based on DE and AD, respectively, and have the same height.

∴S△BDE:S△ABD=DE:AD= 1:3

∫△ABD and△△△ ABC are based on BD and BC, respectively, with the same height.

∴S△ABD:S△ABC=BD:BC=2:5

∴s△bde:s△abc(= 1/3×2/5)=2: 15