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(ΔAPQ))/(A(ΔPQB")) = (square)/(PB)`   ...(i)[areas in proportion of base]

`(A(ΔAPQ))/(A(ΔPQC)) = (square)/(QC)`   ...(ii)[areas in proportion of base]

ΔPQC and ΔPQB have `square` is common base.

Seg PQ || Seg BC, hence height of ΔAPQ and ΔPQB.

A(ΔPQC) = A(Δ`square`)    ......(iii)

`(A(ΔAPQ))/(A(ΔPQB)) = (A(Δsquare))/(A(Δsquare))`   ...[(i), (ii), and (iii)]

`(AP)/(PB) = (AQ)/(QC)`   ...[(i) and (ii)]

Answer 1

Proof: ΔAPQ and ΔPQB have equal height.

`(A(ΔAPQ))/(A(ΔPQB))` = \[\frac{{\boxed{AP}}}{{{PB}}}\]   ...(i)[areas in proportion of base]

`(A(ΔAPQ))/(A(ΔPQC))` = \[\frac{{\boxed{AQ}}}{{{QC}}}\]   ...(ii)[areas in proportion of base]

ΔPQC and ΔPQB have \[\boxed{\text{PQ}}\] is common base.

Seg PQ || Seg BC, hence height of ΔAPQ and ΔPQB.

A(ΔPQC) = A(Δ\[\boxed{\text{PQB}}\])   ...(iii)[Areas of two triangles having equal base and height are equal]

`(A(ΔAPQ))/(A(ΔPQB))` = \[\frac{A(Δ{\boxed{APQ}})}{{A(Δ{\boxed{PQC}}})}\]   ...[(i), (ii), and (iii)]

`(AP)/(PB) = (AQ)/(QC)`   ...[(i) and (ii)]