ABCD is a parallelogram & E is point on CD such that CE=2ED. AE is joined meeting BD in F and BC produced in G. Prove that : AG= 4AF Share with your friends Share 7 Renu Sharma answered this Given that CE= 2ED.so AB= CD= CE + ED= 2ED+ ED= 3ED In ∆DEF and ∆FAB∠FDE = ∠FBA∠DFE =∠BFA∠DEF = ∠FABSo ∆DEF ~ ∆BAFTherefore, DFFB=EFAF=EDABEFAF=ED3ED⇒ EF= AF3 ................. Result (1)Now, In ∆GEC and ∆GABEC is parallel to AB. So both triangles ∆GEC and ∆GAB would be similar.GEGA=GCGB= ECABGEGA= 2ED3EDGE= 2GA3 ................Result (2)Because GE= GA- EA and EA= EF + AFso GE= GA-EF-AF 2GA3= GA-AF3-AF2GA3- GA=-AF3-AF-GA3= -4AF3So GA= 4AF⇒ AG= 4AF Hence Proved. 7 View Full Answer
