Question

In $Fig.6.17$, if $PQRS$ is a parallelogram and $AB\parallel PS$, then prove that $OC\parallel SR$.

Answer 1

Prove the given line parallel

According to the given details

$PQRS$ is a parallelogram

Therefore, $PQ\parallel SR$ and $PS\parallel QR$

Also given that $AB\parallel PS$

We need to prove, $OC\parallel SR$

From $∆OPS$ and $∆OAB$

$$\begin{gathered} \angle POS=\angle AOB\left[\mathrm{Common}\mathrm{angle}\right] \\ \angle OSP=\angle OBA\left[\mathrm{Corresponding}\mathrm{angle}\right] \\ ∆OPS~∆OAB\left[\mathrm{By}\mathrm{AAA}\mathrm{similarity}\mathrm{criteria}\right] \end{gathered}$$

Applying basic proportionality theorem, we obtain

$\frac{PS}{AB}=\frac{OS}{OB}.........................\left(1\right)$

From $∆CQR$ and $∆CAB$

$$\begin{gathered} QR\parallel PS\parallel AB \\ \angle QCR=\angle ACB\left[\mathrm{Common}\mathrm{angle}\right] \\ \angle CQR=\angle CBA\left[\mathrm{Corresponding}\mathrm{angle}\right] \\ ∆CQR~∆CAB \end{gathered}$$

Then, by basic proportionality theorem

$$\begin{gathered} \Rightarrow \frac{QR}{AB}=\frac{CR}{CB} \\ \Rightarrow \frac{PC}{AB}=\frac{CR}{CB}.....................\left(2\right) \end{gathered}$$

From equation $\left(1\right)$ and $\left(2\right)$

$$\begin{gathered} \Rightarrow \frac{OS}{OB}=\frac{CR}{CB} \\ \Rightarrow \frac{OB}{OS}=\frac{CB}{CR} \end{gathered}$$

Subtracting $1$ from $\mathrm{L}.\mathrm{H}.\mathrm{S}$ and $\mathrm{R}.\mathrm{H}.\mathrm{S}$, we get,

$$\begin{gathered} \Rightarrow \frac{OB}{OS}-1=\frac{CB}{CR}-1 \\ \Rightarrow \frac{OB-OS}{OS}=\frac{CB-CR}{CR} \\ \Rightarrow \frac{BS}{OS}=\frac{BR}{CR} \end{gathered}$$

$OC\parallel SR$ by converse of basic property theorem

Hence proved, $OC\parallel SR$.