$E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\Delta PQR$. For the given case,state whether $EF || QR$:
$PE = 4 \, cm, QE = 4.5 \, cm, PF = 8 \, cm$ and $RF = 9 \, cm$.

(N/A) Given:
$PE = 4 \, cm, QE = 4.5 \, cm, PF = 8 \, cm, RF = 9 \, cm$.
According to the Converse of the Basic Proportionality Theorem (Thales Theorem),if a line divides any two sides of a triangle in the same ratio,then the line is parallel to the third side.
Calculate the ratios:
$\frac{PE}{EQ} = \frac{4}{4.5} = \frac{40}{45} = \frac{8}{9}$.
$\frac{PF}{FR} = \frac{8}{9}$.
Since $\frac{PE}{EQ} = \frac{PF}{FR}$,by the Converse of the Basic Proportionality Theorem,$EF$ is parallel to $QR$ $(EF || QR)$.