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Question

# Prove that any line parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally.

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Solution

## Step 1: Note the given dataLet $ABCD$ be a trapezium with $AB\parallel DC$$AD$ and $BC$ are non-parallel sides. Let, $EF$ be a line parallel to $AB,DC$.Join $AC$ such that $△ACD$ and $△ACB$ are triangles and $AC$ meets $EF$ at $G$.Step 2: Apply Thales theorem on $∆ACD$ and $∆ACB$Basic Proportionality Theorem: In a triangle, a line drawn parallel to one side to intersect the other sides distinct points divides two sides in same ratio.In $∆ACD$Here, $EG\parallel DC$. According to Basic Proportionality Theorem$\frac{AE}{ED}=\frac{AG}{GC}$……………..(i)In $∆ACB$Here, $GF\parallel AB$. According to Basic Proportionality Theorem$\begin{array}{l}\frac{FC}{BF}=\frac{GC}{AG}\\ ⇒\frac{BF}{FC}=\frac{AG}{GC}......................\left(ii\right)\end{array}$Equating equation (i) and equation (ii)$\frac{AE}{ED}=\frac{BF}{FC}$Hence proved that a line drawn parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally.

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