Question

In the given figure, X is any point in the interior of triangle. Point X is joined to vertices of triangle. Seg PQ || seg DE, seg QR || seg EF. Fill in the blanks to prove that, seg PR || seg DF.

Answer 1

Given:
Seg PQ || seg DE
seg QR || seg EF
In △DXE, PQ || DE
$\frac{\mathrm{XP}}{\boxed{\mathrm{PD}}}=\frac{\boxed{\mathrm{XQ}}}{\mathrm{QE}}...\left(\mathrm{I}\right)\left(\mathrm{By}\mathrm{basic}\mathrm{proportionality}\mathrm{theorem}\right)$
In △XEF, QR || EF ....Given
$\therefore \frac{\boxed{\mathrm{XQ}}}{\boxed{\mathrm{QE}}}=\frac{\boxed{\mathrm{XR}}}{\boxed{\mathrm{RF}}}.....\left(\mathrm{II}\right)\left(\mathrm{By}\mathrm{basic}\mathrm{proportionality}\mathrm{theorem}\right)$
$\therefore \frac{\boxed{\mathrm{XP}}}{\boxed{\mathrm{PD}}}=\frac{\boxed{\mathrm{XR}}}{\boxed{\mathrm{RF}}}\mathrm{From}\left(\mathrm{I}\right)\mathrm{and}\left(\mathrm{II}\right)$
∴ seg PR || seg DF (Converse of basic proportional theorem)