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प्रश्न
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)
उत्तर
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)
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