Question

Prove that 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

Given: In ΔABC, a line DE || BC intersects AB at D and AC at E.

To Prove: `(AD)/(DB) = (AE)/(EC)`.

Construction: Join BE and CD. Draw DM ⊥ AC and EN ⊥ AB.

Proof:

Area (ΔADE) = `1/2 xx "base"  AD xx "height"  EN`.

Area (ΔBDE) = `1/2 xx "base"  DB xx "height"  EN`.

Therefore, `("Area"(ΔADE))/("Area"(ΔBDE)) = (AD)/(DB)`   ...(1)

Similarly:

Area (ΔADE) = `1/2 xx "base"  AE xx "height"  DM`.

Area (ΔCED) = `1/2 xx "base"  EC xx "height"  DM`.

Therefore, `("Area"(ΔADE))/("Area"(ΔCED)) = (AE)/(EC)`   ...(2)

Now, ΔBDE and ΔCED are on the same base DE and between the same parallels DE || BC.

So, Area (ΔBDE) = Area (ΔCED)   ...(3)

From (1), (2) and (3), we get:

`(AD)/(DB) = (AE)/(EC)`

The ratio of the segments is equal.

Hence Proved.