Question

E and F are points on the side PQ and PR respectively of $△$PQR.For the following case , state whether EF$\parallel$QR
PE=4 cm , QE=4.5 cm , PF =8 cm and RF= 9 cm

Answer 1

Here,
PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm



$\frac{\mathrm{PE}}{\mathrm{QE}}=\frac{4}{4.5}=\frac{8}{9}$

$$\begin{gathered} \frac{\mathrm{PF}}{\mathrm{FR}}=\frac{8}{9} \\ \therefore \frac{\mathrm{PE}}{\mathrm{QE}}=\frac{\mathrm{PF}}{\mathrm{FR}} \end{gathered}$$

According to the converse of basic proportionality theorem, if a line divides any two sides of triangle in the same ratio, then the line is parallel to the third side.

Here,

$\frac{\mathrm{PE}}{\mathrm{QE}}=\frac{\mathrm{PF}}{\mathrm{FR}}$

$\therefore \mathrm{EF}\parallel \mathrm{QR}$