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. 

Proof :  In ΔXDE, PQ || DE         ...`square`

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

In ΔXEF, QR || EF                       ...`square`

∴ `square/square = square/square                                                           ..."(II)" square`

∴ `square/square = square/square`                                ...from (I) and (II)

∴ seg PR || seg DF           ...(converse of basic proportionality theorem)

Answer 1

Given:
Seg PQ || seg DE
seg QR || seg EF

To Prove: seg PR || seg DF

Proof :  

In ΔXDE, PQ || DE         ... Given

∴ `"XP"/underline("PD") = underline("XQ")/"QE"`                ...(I)(Basic proportionality theorem)

In ΔXEF, QR || EF           ... Given

∴ `underline("XR")/underline("RF") = underline("XQ")/underline("QE")`                 ...(II)(Basic proportionality theorem)

∴ `underline("XP")/underline("PD") = underline("XR")/underline("RF")`                  ...from (I) and (II)

∴ seg PR || seg DF           ...(converse of basic proportionality theorem)