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State “Basic Proportionality Theorem” and use it to prove the following :
A line through the mid-point of one side of a triangle, parallel to another side, bisects the third side.
A line through the mid-point of one side of a triangle, parallel to another side, bisects the third side.
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Solution : Statement : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Given : In $\Delta ABC$, $P$ is mid-point of $AB$ and $PQ \parallel BC$
To prove : $Q$ is the mid-point of $AC$
Proof : As $PQ \parallel BC$
$\frac{AP}{PB} = \frac{AQ}{QC}$ (by BPT)
$AP = PB$ (given) $\implies \frac{AP}{PB} = 1$
$\therefore \frac{AQ}{QC} = 1 \implies AQ = QC$
$\implies Q$ is the mid-point of $AC$
Given : In $\Delta ABC$, $P$ is mid-point of $AB$ and $PQ \parallel BC$
To prove : $Q$ is the mid-point of $AC$
Proof : As $PQ \parallel BC$
$\frac{AP}{PB} = \frac{AQ}{QC}$ (by BPT)
$AP = PB$ (given) $\implies \frac{AP}{PB} = 1$
$\therefore \frac{AQ}{QC} = 1 \implies AQ = QC$
$\implies Q$ is the mid-point of $AC$