Question

​ In a cyclic quadrilateral ABCD , E is a point on AC such that angle ADE = angle BDC. PT AD.BC= AE. DB and AB.DE=CE. DA

Answer 1

see the diagram.

In a cyclic quadrilateral, the vertices are on the circumference of a circle. Also, the angle between a side and a diagonal is equal to the angle between the opposite side and the other diagonal.

Given  ∠ADE = ∠CDB = x (let). 
ABCD is a Cyclic Quadrilateral
   =>  ∠CAD = ∠CBD  = y      and     ∠CDB = ∠CAB = x
   =>  ∠ABD = ACD = w   
Also,  ∠CDE = x + z  = ∠ADB

Compare ΔAED and ΔBCD.  The angles are same and hence are similar triangles.  Ratios of corresponding sides are:
  =>   AD/ BD = AE / BC
  =>   AD * BC  = AE * BD      proved.

Now,  compare ΔABD  and ΔCDE.  Two angles x+z and w  are equal in both. They are similar triangles. Hence,
  =>  AB / DA =   CE / DE
  =>  AB * DE = CE * DA      proved.