E and F are points on the sides PQ and PR res | vedclass

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(A) Given: $PQ = 1.28 \, cm, PR = 2.56 \, cm, PE = 0.18 \, cm, PF = 0.36 \, cm$.According to the converse of the Basic Proportionality Theorem (Thales

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(A) Given: $PQ = 1.28 \, cm, PR = 2.56 \, cm, PE = 0.18 \, cm, PF = 0.36 \, cm$.
According to the converse of the Basic Proportionality Theorem (Thales Theorem),if a line divides two sides of a triangle in the same ratio,then the line is parallel to the third side.
Calculate the ratios:
$\frac{PE}{PQ} = \frac{0.18}{1.28} = \frac{18}{128} = \frac{9}{64}$
$\frac{PF}{PR} = \frac{0.36}{2.56} = \frac{36}{256} = \frac{9}{64}$
Since $\frac{PE}{PQ} = \frac{PF}{PR} = \frac{9}{64}$,the line $EF$ divides the sides $PQ$ and $PR$ in the same ratio.
Therefore,by the converse of the Basic Proportionality Theorem,$EF || QR$.

$E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\Delta PQR$. For each of the following cases,state whether $EF || QR$. $PQ = 1.28 \, cm, PR = 2.56 \, cm, PE = 0.18 \, cm$ and $PF = 0.36 \, cm$.

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