Question

$can \: ya \: all \: please \: help$
D and E are points on the sides AB and AC respectively of ∆ABC such that DE is parallel to BC, and AD:DB=4:5. CD and BE intersect each other at F.Find the ratio of the areas of ∆DEF and ∆CBF.

Answer 1

DE || BC 
AD/DB = AE/EF =4/5 ( Basic proportionality theorem) 
In triangle ABE & ACD 

AD/DB = AE/EC and angle A is common 
So triangles ABE & ACD are similar 
Therefore AD/DB = EF/FB ( properties of similar triangles)
Similarly AE / EC = BF/FC 
In Triangles AEF & BFC 
angle FEC is congruent to angle BFE ( alternate angles)
angel DEF is = angle BFC ( vertically opposite angles)
Therefore they are similar 
Area of triangle DFE / area of BFC = (4/5)^2 ( Properties of similar triangles 
= 16/25