In figure, A, B and C are points on OP,


In figure, $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are points on $\mathrm{OP}, \mathrm{OQ}$ and $\mathrm{OR}$ respectively such that $\mathrm{AB} \| \mathrm{PQ}$ and $\mathrm{AC}$ $\| \mathrm{PR}$. Show that $\mathrm{BC} \| \mathrm{QR}$.


In $\triangle \mathrm{POO}$

AB $\| P Q$ (given)

$\frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OA}}{\mathrm{AP}} \ldots$ (i) (Basic Proportionality Theorem)

In $\triangle \mathrm{POR}$

$\mathrm{AC} \| \mathrm{PR}$ (given)

$\frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OC}}{\mathrm{CR}} \ldots$ (ii) (Basic Proportionality Theorem)

From (i) and (ii), we get


$\therefore$ By converse of Basic Proportionality Theorem,

$B C \| O R$

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