Question

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

Answer 1

We are given a triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively.

We need to prove that AD/ AE = DB /EC.

Let us join BE and CD and then draw DM$\perp$AC and EN$\perp$AB.

Now, area of $\Delta$ADE ( 1/ 2 base$\times$height) = 1/ 2 AD$\times$ EN. area of $\Delta$ADE is denoted as ar(ADE).

So, ar(ADE) = 1/ 2 AD$\times$EN

Similarly, ar(BDE) = 1 /2 DB$\times$EN, ar(ADE) = 1/ 2 AE$\times$DM and ar(DEC) =1 /2 EC$\times$DM.

Therefore, ar(ADE)/ ar(BDE) =$\frac{\frac{1}{2}\times \text{AD}\times \text{EN}}{\frac{1}{2}\times \text{DB}\times \text{EN}}=\frac{AD}{DB}$ --------(1)

ar(ADE) / ar(DEC) = $\frac{\frac{1}{2}\times \text{AE}\times \text{DM}}{\frac{1}{2}\times \text{EC}\times \text{DM}}=\frac{AE}{EC}$ -------------(2)

Note that Δ BDE and Δ DEC are on the same base DE and between the same parallels BC and DE.

So, ar(BDE) = ar(DEC) ------------(3)

Therefore, from (1), (2) and (3), we have:

AD/ DB = AE/ EC