Solution

(i)

Let $EC=xcm$

It is given that $DE\Vert BC$.

By using basic proportionality theorem, we obtain

$\begin{array}{rl}& {\displaystyle \frac{AD}{DB}}={\displaystyle \frac{AE}{EC}}\\ & {\displaystyle \frac{1.5}{3}}={\displaystyle \frac{1}{x}}\\ & x={\displaystyle \frac{3\times 1}{1.5}}\\ & x=2\\ & \therefore EC=2cm\end{array}$

(ii)

Let $AD=xcm$

It is given that DE || BC.

By using basic proportionality theorem, we obtain

${\displaystyle \frac{AD}{DB}}={\displaystyle \frac{AE}{EC}}$

${\displaystyle \frac{x}{7.2}}={\displaystyle \frac{1.8}{5.4}}$

$x={\displaystyle \frac{1.8\times 7.2}{5.4}}$

$x=2.4$

$\therefore AD=2.4cm$

Solution

(i)

Given that, $PE=3.9cm,EQ=3cm,PF=3.6cm,FR=2.4cm$ ${\displaystyle \frac{PE}{EQ}}={\displaystyle \frac{3.9}{3}}=1.3$ ${\displaystyle \frac{PF}{FR}}={\displaystyle \frac{3.6}{2.4}}=1.5$

Hence, ${\displaystyle \frac{PE}{EQ}}\ne {\displaystyle \frac{PF}{FR}}$

Therefore, EF is not parallel to QR.

(ii)

$PE=4cm,QE=4.5cm,PF=8cm,RF=9cm$

${\displaystyle \frac{PE}{EQ}}={\displaystyle \frac{4}{4.5}}={\displaystyle \frac{8}{9}}$

${\displaystyle \frac{PF}{FR}}={\displaystyle \frac{8}{9}}$

Hence, ${\displaystyle \frac{PE}{EQ}}={\displaystyle \frac{PF}{FR}}$

Therefore, $EF$ is parallel to QR.

(iii)

${\displaystyle \frac{PE}{PQ}}={\displaystyle \frac{0.18}{1.28}}={\displaystyle \frac{18}{128}}={\displaystyle \frac{9}{64}}$

${\displaystyle \frac{PF}{PR}}={\displaystyle \frac{0.36}{2.56}}={\displaystyle \frac{9}{64}}$

Hence, ${\displaystyle \frac{PE}{PQ}}={\displaystyle \frac{PF}{PR}}$

Therefore, EF is parallel to QR.

Solution

In the given figure, $LM\Vert CB$

By using basic proportionality theorem, we obtain ${\displaystyle \frac{AM}{AB}}={\displaystyle \frac{AL}{AC}}$

Similarly, $LN\Vert CD$

$\therefore {\displaystyle \frac{AN}{AD}}={\displaystyle \frac{AL}{AC}}$

From $(i)$ and $(ii)$, we obtain

${\displaystyle \frac{AM}{AB}}={\displaystyle \frac{AN}{AD}}$

Solution

In $\mathrm{△}ABC,DE\Vert AC$

$\therefore {\displaystyle \frac{BD}{DA}}={\displaystyle \frac{BE}{EC}}$ (Basic Proportionality Theorem)

In $\mathrm{△}BAE,DF\Vert AE$

$\therefore {\displaystyle \frac{BD}{DA}}={\displaystyle \frac{BF}{FE}}$ (Basic Proportionality Theorem)

(ii)

From $(i)$ and $(ii)$, we obtain

${\displaystyle \frac{BE}{EC}}={\displaystyle \frac{BF}{FE}}$

Solution

In $\mathrm{△}POQ,DE\Vert OQ$

$\therefore {\displaystyle \frac{PE}{EQ}}={\displaystyle \frac{PD}{DO}}$ (Basic proportionality theorem)

(i)

In $\mathrm{△}POR,DF\Vert OR$

$\therefore {\displaystyle \frac{PF}{FR}}={\displaystyle \frac{PD}{DO}}$ (Basic proportionality theorem)

From $(i)$ and $(ii)$, we obtain

${\displaystyle \frac{PE}{EQ}}={\displaystyle \frac{PF}{FR}}$

$\therefore EF\Vert QR$ (Converse of basic proportionality theorem)

Solution

In $\mathrm{△}POQ,AB\Vert PQ$

$\therefore {\displaystyle \frac{OA}{AP}}={\displaystyle \frac{OB}{BQ}}$ (Basic proportionality theorem)

(i)

In $\mathrm{△}POR,AC\Vert PR$

$\therefore {\displaystyle \frac{OA}{AP}}={\displaystyle \frac{OC}{CR}}$ (By basic proportionality theorem) (ii)

From $(i)$ and $(ii)$, we obtain

${\displaystyle \frac{OB}{BQ}}={\displaystyle \frac{OC}{CR}}$

$\therefore BC\Vert QR$ (By the converse of basic proportionality theorem)

Solution

Consider the given figure in which $PQ$ is a line segment drawn through the mid-point $P$ of line $AB$, such that $PQ\Vert BC$

By using basic proportionality theorem, we obtain

${\displaystyle \frac{AQ}{QC}}={\displaystyle \frac{AP}{PB}}$

${\displaystyle \frac{AQ}{QC}}={\displaystyle \frac{1}{1}}$ (P is the mid-point of $AB.\therefore AP=PB$ )

$\Rightarrow AQ=QC$

Or, $Q$ is the mid-point of $AC$.

Solution

Consider the given figure in which $PQ$ is a line segment joining the mid-points $P$ and $Q$ of line $AB$ and $AC$ respectively.

i.e., $AP=PB$ and $AQ=QC$

It can be observed that ${\displaystyle \frac{AP}{PB}}={\displaystyle \frac{1}{1}}$

and ${\displaystyle \frac{AQ}{QC}}={\displaystyle \frac{1}{1}}$

$\therefore {\displaystyle \frac{AP}{PB}}={\displaystyle \frac{AQ}{QC}}$

Hence, by using basic proportionality theorem, we obtain

$PQ\Vert BC$

Solution

Draw a line $EF$ through point $O$, such that

$EF\Vert CD$

In $\mathrm{△}ADC,EO\Vert CD$

By using basic proportionality theorem, we obtain

${\displaystyle \frac{AE}{ED}}={\displaystyle \frac{AO}{OC}}$

In $\mathrm{△}ABD$

$OE\Vert AB$

So, by using basic proportionality theorem, we obtain

${\displaystyle \frac{ED}{AE}}={\displaystyle \frac{OD}{BO}}$

$\Rightarrow {\displaystyle \frac{AE}{ED}}={\displaystyle \frac{BO}{OD}}$

From equations (1) and (2), we obtain ${\displaystyle \frac{AO}{OC}}={\displaystyle \frac{BO}{OD}}$

$\Rightarrow {\displaystyle \frac{AO}{BO}}={\displaystyle \frac{OC}{OD}}$

Solution

Let us consider the following figure for the given question.

Draw a line $OE\Vert AB$

In $\mathrm{△}ABD,OE\Vert AB$

By using basic proportionality theorem, we obtain

${\displaystyle \frac{AE}{ED}}={\displaystyle \frac{BO}{OD}}$

However, it is given that ${\displaystyle \frac{AO}{OC}}={\displaystyle \frac{OB}{OD}}$

From equations (1) and (2), we obtain

${\displaystyle \frac{AE}{ED}}={\displaystyle \frac{AO}{OC}}$

$\Rightarrow EO$ || DC [By the converse of basic proportionality theorem]

$\Rightarrow AB||OE||DC$

$\Rightarrow AB\Vert CD$

$\therefore ABCD$ is a trapezium.