Question

triangle ABC and triangle EBC on same base BC.If AE produced intersect BC atD .prove
ar (ABC)/ar (EBC)=AD/ED

Answer 1

See the diagram.
 
There are two possibilities.  Draw the perpendicular from A in case 1.
  Draw perpendiculars from A and E in case 2.  

Case 1:
   GF is equal to altitude of triangle EBC. Also the triangles AGE and AFD are similar. It is clear as all sides are parallel to corresponding sides.

So AE / AD = AG / AF
     1 -  AE / AD  =  1 -    AG / AF
      ED / AD =  GF / AF   -- -(1)

Ar(ABC) / Ar(EBC)
   = 1/2 BC * AF / [1/2 * BC * GF
   = AF / GF
   =  AD / ED      from (1)
========

Case 2:  simple

Ar (ABC) / Ar(EBC)
  = 1/2 * BC * AF  /  [1/2 * BC * EG ]
  = AF / EG
  = AD / ED      as  the right angle triangles  AFD and EGD are similar ,
                               ∵ ∠ADF = ∠EDG   vertical angles.
                                   ∠AFD = ∠EGD