Question

In a trapezium ABCD AB parallel to CD and DC=3AB EF parallel toAB intersect DA and CB respectively at E andF such that BFdivided byFC=2by3.prove that 3DC=5EF

Answer 1

See the diagram.    We use similar Δs.

ΔAJD and  ΔAGE are similar.   AG / AJ = GE / JD   --(1)

BF / BC = 2 : (2+3) = 2/5

ΔBHF and ΔBIC are similar.
    HF / IC = BH / BI = BF / BC  = 2 : 5  --(2)

Quadrilaterals AGHB and AJIB are rectangles as the corresponding sides  are || and internal angles are 90°. 
So AG / AJ = BH/BI
Hence from (1) and (2) we get:

GE / JD = HF / IC = 2 / 5     or,   JD = 2.5 GE   &   IC = 2.5 HF

Now   DC = DJ + JI + IC
=>   3 AB  = 2.5 GE + AB + 2.5 HF 
=>   4 AB / 5 = GE + HF
=>   EF = GH + GE + HF
             = AB + 4 AB/ 5 = 9 AB / 5 
=>   EF = 3 DC/  5

=>   5 EF = 3 DC