Question

Prove that, if a line parallel to a side of a triangle intersects the other sides in two district points, then the line divides those sides in proportion.

Prove that, 'If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the side in the same proportion’.

Answer 1

Given: A Δ ABC in which DE||BC, and intersect AB in D and AC in E.

To Prove: `"AD"/"BD" = "AE"/"EC"`

Construction: Join BE, CD, and draw EF⊥ BA and DG⊥CA.

Proof:

Area ∆ADE = `1/2 xx ("base" xx "height") =1/2("AD. EF")`

Area ∆DBE = `1/2 xx("base" xx "height") =1/2("DB.EF")`

⇒  `("Area"(triangle"ADE"))/("Area"(triangle"DBE"))=[((1/2"AD. EF"))/((1/2"DB. EF")]] = "AD"/"DB"` 

Similarly,

⇒ `("Area"(triangle"ADE"))/("Area"(triangle"DBE"))=[((1/2"AE. DG"))/((1/2"EC. DG")]] = "AE"/"EC"` 

ΔDBE and ΔDEC are on the same base DE and BC the same parallel DE and BC.

∴ Area (ΔDBE) = Area (ΔDEC)

⇒ `1/("Area"(triangle"DBE")) =1/("Area"(triangle"DEC"))`   ...[Taking reciprocal of both sides]

⇒ `("Area"(triangle"ADE"))/("Area"(triangle"DBE"))=("Area"(triangle"ADE"))/("Area"(triangle"DEC")`   ...[Multiplying both sides by Area (ΔADE)]

⇒ `"AD"/"DB"="AE"/"EC"`

Hence Proved.

Answer 2

Given: In Δ ABC line l || line BC and line l intersects AB and AC in point P and Q respectively.

To prove: `"AP"/"PB" = "AQ"/"QC"`

Construction: Draw seg PC and seg BQ.

Proof: ΔAPQ and ΔPQB have equal heights.

`therefore ("A"(Delta "APQ"))/("A"(Delta "PQB")) = "AP"/"PB"`    (areas proportionate to bases)   ...(I)

and `("A"(Delta "APQ"))/("A"(Delta "PQC")) = "AQ"/"QC"`   (areas proportionate to bases)   ...(II)

Seg PQ is the common base of ΔPQB and ΔPQC, seg PQ || seg BC,

Hence ΔPQB and ΔPQC have equal areas.

A(ΔPQB) = A(ΔPQC)   ...(III)

`("A"(Delta "APQ"))/("A"(Delta "PQB")) = ("A"(Delta "APQ"))/("A"(Delta "PQC"))`    ...[from (I), (II) and (III)]

`therefore "AP"/"PB" = "AQ"/"QC"`   ...[from (I) and (II)]

Hence Proved.