State basic proportionality theorem.
Use it to prove the following:
If three parallel lines $l$, $m$, $n$ are intersected by transversals $q$ and $s$ as shown in the adjoining figure, then $\frac{AB}{

Question

State basic proportionality theorem.
Use it to prove the following:
If three parallel lines $l$, $m$, $n$ are intersected by transversals $q$ and $s$ as shown in the adjoining figure, then $\frac{AB}{BC} = \frac{DE}{EF}$.

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Accepted Answer

Step 1: **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.
Step 2: Join $AF$ intersecting line $m$ at $G$.
Step 3: In $	riangle ACF$, $BG \parallel CF$
$\Rightarrow \frac{AB}{BC} = \frac{AG}{GF}$ ... (i) (by BPT)
Step 4: In $	riangle FDA$, $GE \parallel AD$
$\Rightarrow \frac{EF}{DE} = \frac{GF}{AG} \Rightarrow \frac{DE}{EF} = \frac{AG}{GF}$ ... (ii) (by BPT)
Step 5: From (i) and (ii), we get $\frac{AB}{BC} = \frac{DE}{EF}$.

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