Question

plz answer fast it's urgent

Answer 1

Proof- AB = DC [Sides of parallelogram]       .........................(1)
 
    In triangles. DCE and FBE
               ∠E =  ∠E     [Vertically opp. ∠s]
        and, ∠D = ∠F     [Alt. Int ∠s]

so, Triangle DCE & FBE are Similar By AA Criteria

hence, DC/FB = CE/BE = DE/FE

But CE/BE = 1 [∵E is the mid point]

so, 1 = DC/FB
 so, DC =FB

so By Eqn (1)

AB= FB
 so, AF = AB+FB
      AF = 2AB                 [∵AB=FB]