Question

Prove that If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. In the figure, find EC if `(AD)/(DB) = (AE)/(EC)` using the above theorem.

Answer 1

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

We need to prove that `(AD)/(DB) = (AE)/(EC)`

Let us join BE and CD and then draw DM ⊥ AC and EN ⊥ AB.

Now, area of ∆ADE = `1/2` (base × height) = `1/2` AD × EN

The area of ∆ADE is denoted as ar(ADE).

So, ar(ADE) = `1/2` AD × EN

Similarly, ar(BDE) = `1/2` DB × EN,

ar(ADE) = `1/2` AE × DM and ar(DEC) = `1/2` EC × DM

 Therefore, `(ar(ADE))/(ar(BDE)) = (1/2 AD xx EN)/(1/2 DB xx EN) = (AD)/(DB)`  ......(1)

And `(ar(ADE))/(ar(DEC)) = (1/2 AE xx DM)/(1/2 EC xx DM) = (AE)/(EC)`  ......(2)

Note that ∆BDE and 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)`