Question

In the figure, X is any point in the interior of the triangle. Point X is joined to vertices of triangle. seg PQ || seg DE, seg QR || seg EF. 
Complete the following activity to prove seg PR || seg DF.

Activity:

In ΔXDE, PQ || DE     ...(given)

∴ `XP/square = square/"QE"`    ...(I) Basic proportionality theorem

In ΔXEF, QR || EF ......(Given)

∴ `"XQ"/"QE" = square/"RF"`   ...(II)`square`

∴ `"XP"/"PD" = square/square`    ...From (I) and (II)

∴ seg PR || seg DF     ....Converse of basic proportionality theorem

Answer 1

In ΔXDE, PQ || DE     ...(given)

\[{\frac{\text{XP}}{\boxed{\text{PD}}} = \frac{\boxed{\text{XO}}}{\text{QE}}}\]    ...(I) Basic proportionality theorem

In ΔXEF, QR || EF ......(Given)

∴ `"XQ"/"QE"` = \[\frac{\boxed{\text{XR}}}{\text{RF}}\]   ...(II)\[\boxed{\text{B.P.T.}}\]

∴ `"XP"/"PD"`= \[\frac{\boxed{\text{XR}}}{\boxed{\text{RF}}}\]    ...From (I) and (II)

∴ seg PR || seg DF     ....Converse of basic proportionality theorem