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Question
A line is parallel to one side of triangle which intersects remaining two sides in two distinct points then that line divides sides in same proportion.
Given: In ΔABC line l || side BC and line l intersect side AB in P and side AC in Q.
To prove: `"AP"/"PB" = "AQ"/"QC"`
Construction: Draw CP and BQ
Proof: ΔAPQ and ΔPQB have equal height.
`("A"(Δ"APQ"))/("A"(Δ"PQB")) = (["______"])/"PB"` .....(i)[areas in proportion of base]
`("A"(Δ"APQ"))/("A"(Δ"PQC")) = (["______"])/"QC"` .......(ii)[areas in proportion of base]
ΔPQC and ΔPQB have [______] is common base.
Seg PQ || Seg BC, hence height of ΔAPQ and ΔPQB.
A(ΔPQC) = A(Δ______) ......(iii)
`("A"(Δ"APQ"))/("A"(Δ"PQB")) = ("A"(Δ "______"))/("A"(Δ "______"))` ......[(i), (ii), and (iii)]
`"AP"/"PB" = "AQ"/"QC"` .......[(i) and (ii)]
Solution
Proof: ΔAPQ and ΔPQB have equal height.
`("A"(Δ"APQ"))/("A"(Δ"PQB"))` = `"AP"/"PB"` .....(i)[areas in proportion of base]
`("A"(Δ"APQ"))/("A"(Δ"PQC"))` = `"AQ"/"QC"` .......(ii)[areas in proportion of base]
ΔPQC and ΔPQB have PQ is common base.
Seg PQ || Seg BC, hence height of ΔAPQ and ΔPQB.
A(ΔPQC) = A(ΔPQB) ......(iii)[Areas of two triangles having equal base and height are equal]
`("A"(Δ"APQ"))/("A"(Δ"PQB"))` = `("A"(Δ"APQ"))/("A"(Δ "PQC"))` ......[(i), (ii), and (iii)]
`"AP"/"PB" = "AQ"/"QC"` .......[(i) and (ii)]
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