Question

if a line divides any two sides of a triangle in the same ratio,then the line is parallel to the third side<br />prove it

Answer 1

In ΔABC , let AB and AC be divided by a line DE in equal proportions.

So  AD / DB = AE / EC.

So   DB/AD = EC / AE

****  Add 1 on both sides.

      (DB+AD)/AD = (EC+AE)/AE 
       AB/ AD = AC / AE
=>   AD/ AB = AE / AC

In the two Δ ADE and ABC, angle ∠A is the same. AD || AB. AE || AC. And the sides are proportional.

    Hence both triangles are similar.

        So  DE || BC.              Proved.

Answer 2

Hey there,

This is one method, but there are two ways to do it.

The line DE divides AB and AC at D and E such that
AD / DB = AE / EC
According to theorem the lines are parallel.TPT:  DE || BC,  Assume DE is not parallel to BC.Then construct DF||BCThen by theorem 6.1, 
AE / EC = AF / FC
But both points E and F are on AC dividing AC in the same ratio.  So E and F are just one point and hence the result and our assumption is wrong.  ThereforeDE || BC.

Hope this helps!