Question

seg ED$\parallel$seg QO and
seg DF$\parallel$seg OR
Prove that seg EF$\parallel$side QR

Answer 1

Given: seg ED$\parallel$seg QO
seg DF$\parallel$seg OR

To prove: seg EF$\parallel$side QR

Proof:
$$\begin{gathered} \mathrm{In}△\mathrm{POQ} \\ \mathrm{seg}\mathrm{ED}\parallel \mathrm{seg}\mathrm{QO} \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{basic}\mathrm{proportionality}\mathrm{theorem}, \\ \frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PD}}{\mathrm{DO}}...\left(1\right) \\ \mathrm{Now}, \\ \mathrm{In}△\mathrm{POR} \\ \mathrm{seg}\mathrm{DF}\parallel \mathrm{seg}\mathrm{OR} \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{basic}\mathrm{proportionality}\mathrm{theorem}, \\ \frac{\mathrm{PD}}{\mathrm{DO}}=\frac{\mathrm{PF}}{\mathrm{FR}}...\left(2\right) \\ \mathrm{From}\left(1\right)\&\left(2\right),\mathrm{we}\mathrm{get}: \\ \frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}} \\ \mathrm{Thus},\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{converse}\mathrm{of}\mathrm{basic}\mathrm{proportionality}\mathrm{theorem},\mathrm{seg}\mathrm{EF}\parallel \mathrm{side}\mathrm{QR}. \end{gathered}$$