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Questions
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’.
Solution 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.
Solution 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.
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