Question

Prove that a line drawn parallel to one side of a triangle intersecting the other two sides in distinct points divides the other two sides in the same ratio.

Answer 1

Given: In ∆ABC, a line (l) drawn parallel to side BC intersects AB and AC at D and E, respectively.

To prove: `(AD)/(DB) = (AE)/(EC)`

Construction: Draw perpendiculars from D and E to AC and AB i.e., DM ⊥ AC and EN ⊥ AB.

Join DC and BE.

Proof:

`(ar(ΔADE))/(ar(ΔBDE))`

= `(1/2(AD)(EN))/(1/2(BD)(EN))`

= `(AD)/(DB)`   ...(1)

`(ar(ΔADE))/(ar(ΔCED))`

= `(1/2(AE)(DM))/(1/2(EC)(DM))`

= `(AE)/(EC)`   ...(2)

Also, ar(ΔBDE) = ar(ΔCED)   ...(3) (Triangles on same base and between same parallel are equal in area.)

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

`(ar(ΔADE))/(ar(ΔBDE)) = (ar(ΔADE))/(ar(ΔCED))`

⇒ `(AD)/(DB) = (AE)/(EC)`

Hence proved.