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)]

Answer 1

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)]