Answers

  • 369
  • Ambitious
2015-07-31T14:05:13+05:30

Given
    In ΔABC, D and E are the two points of AB and AC respectively, 
    such that, AD/DB = AE/EC. 

To Prove
    DE || BC 

Proof
    In ΔABC,
     given, AD/DB = AE/EC ----- (1)

    Let us assume that in ΔABC, the point F is an intersect on the side AC.
     So we can apply the Thales Theorem, AD/DB = AF/FC ----- (2)

    Simplify, in (1) and (2) ==> AE/EC = AF/FC 
    Add 1 on both sides, ==> (AE/EC) + 1 = (AF/FC) + 1 
    ==> (AE+EC)/EC = (AF+FC)/FC 
    ==> AC/EC = AC/FC 
    ==> EC = FC 
    From the above, we can say that the points E and F coincide on AC. 
    i.e., DF coincides with DE. 

    Since DF is parallel to BC, DE is also parallel BC  prove
0