Question

please solve question number 24

Answer 1

NOW, in ∆BCG , DE //GC

BD =BE =1

DG EC 1

therefore,BE=BD AND DG=EC

GIVEN: 2EC=BE

EC=GD

GD=1/2BE

PUT THIS VALUE IN 1 WE GET :

AD =AF

1/2BE CF

2AD= AF

BE CF

HENCE: 2AD×CF=AF×BE

PROVED