Question

/_ BED = /_ BRE and E divides BC in the ratio 2:1
Prove that AF × BE = 2 AD × CF

Answer 1

Answer:

Step-by-step explanation:

Through C draw CG∥FD.Now, in ∆AFD, since CG∥FD, thenACCF = AGGD [BPT]⇒ACCF + 1= AGGD + 1⇒AC + CFCF = AG + GDGD⇒AFCF = ADGD .........(1)In ∆BCG, since DE∥GC, thenBEEC = BDGD [BPT]21 = BDGD [as, BE : EC = 2 : 1]BD = 2 GD .......(2)In ∆BDE, we have ∠BED = ∠BDE (given)⇒BD = BE (sides opposite to equal angles are equal) ......(3)From (2) and (3), we get2 GD = BE ⇒GD = BE2 ........(4)Substituting the value of GD from (4) in (1), we getAFCF = ADBE/2⇒AFCF =2ADBE⇒AF × BE = 2 AD × CF