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Question
Prove that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.
Using the above theorem prove that a line through the point of intersection of the diagonals and parallel to the base of the trapezium divides the non-parallel sides in the same ratio.
Solution
For the Theorem:
Given, To prove, Construction and figure
Proof
Let ABCD be a trapezium DC ∥ AB and EF is a line parallel to AB and hence to DC.
To prove: `(DE)/(EA) = (CF)/(FB)`
Construction: Join AC, meeting EF in G.
Proof: In ΔABC, we have
GF || AB
`(CG)/(GA) = (CF)/(FB)` [By BPT] ......(1)
In ΔADC, we have
EG ∥ DC .....(EF ∥AB and AB ∥ DC)
`(DE)/(EA) = (CG)/(GA)` [By BPT] ......(2)
From (1) and (2), we get,
`(DE)/(EA) = (CF)/(FB)`
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