Question

In the figure, points A, B and C are on seg OP, seg OQ and seg OR, respectively, such that AB $\parallel$ PQ and AC $\parallel$ PR. Show that BC $\parallel$QR.

Answer 1

Given : AB $\parallel$ PQ and AC $\parallel$ PR.
To prove:BC $\parallel$QR.

Proof :
$$\begin{gathered} \text{In}△\mathrm{OPQ}, \\ \mathrm{AB}\parallel \mathrm{PQ}. \\ \mathrm{Therefore},\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{basic}\mathrm{proportionality}\mathrm{theorem}, \\ \frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OB}}{\mathrm{BQ}}...\left(1\right) \\ \text{In}△\mathrm{OPR}, \\ \mathrm{AC}\parallel \mathrm{PR}. \\ \mathrm{Therefore},\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{basic}\mathrm{proportionality}\mathrm{theorem}, \\ \frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OC}}{\mathrm{CR}}...\left(2\right) \end{gathered}$$

$$\begin{gathered} \mathrm{From}\left(1\right)\&\left(2\right),\text{we have:} \\ \frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OC}}{\mathrm{CR}} \\ \text{Now},\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{converse}\mathrm{of}\mathrm{basic}\mathrm{proportionality}\mathrm{theorem}, \\ \mathrm{BC}\parallel \mathrm{QR}. \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$