Basic Proportionality Theorem (BPT): If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
Proof that AB/BC = DE/EF:
Join A to F. Draw a line AF intersecting line m at point G (i.e., AG || CF if needed — alternatively, use the standard construction below).
Construction: Join B to D and B to F. Draw BE parallel to CF...
(Standard proof using BPT:)
Draw line segment AD. Since $l \| m \| n$, consider transversal $q$ and diagonal $AF$ as a transversal cutting through a triangle formed by the parallel lines.
Proper Construction: Join A to F. Let AF intersect line $m$ at point G.
In $\triangle$ ACF:
Wait — standard approach:
In $\triangle$ ACF, since $AB \| m$ meets…
Correct proof:
Join A to F. Let AF intersect line $m$ at point G.
In $\triangle$ ACF:
Restarting with correct standard proof:
Join A to F, intersecting line $m$ at G.
In $\triangle$ ACF, $B$ is on AC and $G$ is on AF such that $BG \| CF$ (since $l \| n$ and both are cut by transversals).
By BPT in $\triangle$ ACF:
$$\frac{AB}{BC} = \frac{AG}{GF} \quad \cdots (1)$$
In $\triangle$ DAF, $E$ is on DF and $G$ is on AF such that $EG \| DA$ (since $m \| l$).
By BPT in $\triangle$ DAF:
$$\frac{GE}{...}$$
Final clean proof:
Construction: Join A to F; let it intersect line $m$ at point G.
In $\triangle$ ACF: Since $l \| n$, we have $BG \| CF$.
By BPT: $\dfrac{AB}{BC} = \dfrac{AG}{GF}$ $\quad$ ...(1)
In $\triangle$ DAF: Since $l \| m$, we have $DG \| ... $
Hmm, let me write the definitive version:
Construction: Join A to F. Let AF cut line $m$ at G.
In $\triangle$ CAF: $B$ is on CA, $G$ is on FA.
Since $l \| n$ (given), $BG \| CF$.
By BPT: $\dfrac{AB}{BC} = \dfrac{AG}{GF}$ ...(1)
In $\triangle$ ADF: $G$ is on AF, $E$ is on DF.
Since $m \| l$, $GE \| AD$.
By BPT: $\dfrac{GF}{AG} = \dfrac{EF}{DE}$, i.e., $\dfrac{AG}{GF} = \dfrac{DE}{EF}$ ...(2)
From (1) and (2):
$$\boxed{\dfrac{AB}{BC} = \dfrac{DE}{EF}}$$
Source: Triangles, Section 6.3 (Basic Proportionality Theorem)
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