Question

Points $E$ and $F$ lie on sides $PQ$ and $PR$ of any $△PQR$. From the following, for each case find is $EF\parallel QR$
(i) $PE=3.9cm,EQ=3cm.PF=3.6cm$ and $FR=2.4cm$
(ii) $PE=4cm,QE=4cm,PF=8cm$ and $RF=9cm$
(iii) $PQ=1.28cm,PR=2.56cm,PE=0.18cm$ and $PF=0.36cm$

Answer 1

In $△PQR$, two point $E$ and $F$ lie on sides $PQ$ and $PR$ respectively
(i) $\frac{PE}{EQ}=\frac{3.9}{2}=1.3cm$
$\frac{PF}{FR}=\frac{3.6}{2.4}=\frac{36}{24}=\frac{3}{2}=1.5cm$
$\frac{PE}{EQ}\ne \frac{PF}{FR}$
$EF$ is not parallel to $QR$
(ii) $PE=4cm$, $QE=4.5cm$, $PF=8cm$ and $RF=9cm$
$\frac{PE}{QE}=\frac{4}{4.5}=\frac{40}{45}=\frac{8}{9}$...(i)
and $\frac{PF}{RF}=\frac{8}{9}$
From equation (i) and (ii)
$\frac{PE}{QE}=\frac{PF}{RF}$ (by converse of basic prop. theorem)
$EF\parallel QR$
(iii) $PQ=1.28cm,PR=2.56cm,PE=0.18cm$ and $PF=0.36cm$
$=1.28-0.18=1.10cm$
$FR=PR-PF$
$2.56-0.36=2.20cm$
$\frac{PE}{EQ}=\frac{0.18}{1.10}=\frac{18}{110}=\frac{9}{55}$...(i)
and $\frac{PF}{FR}=\frac{0.36}{2.20}=\frac{36}{220}=\frac{9}{55}$...(ii)
from equation (i) and (ii)
$\frac{PE}{EQ}=\frac{PF}{FR}$ (by converse of basic prop. theorem)
$EF\parallel QR$