Question:

In the figure 1.44, 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 XDE, PQ || DE ……..______

XP/_______ = ______/ QE ………. (I) (Basic proportionality theorem)

In XEF, QR || EF ………. _______

______ /______= ______ /______ ……….(II) ______

______ /______= ______ /______ ………. from (I) and (II)

seg PR || seg DE ………. (converse of basic proportionality theorem)

Answer:

In XDE, PQ || DE……. Given
XP/PD = XQ/QE…….. (I) (Basic proportionality theorem)
In XEF, QR || EF……. Given
XR/RF = XQ/QE……. .(II) (Basic proportionality theorem)
XP/PD = XR/RF ………. from (I) and (II)
seg PR || seg DE ………. (converse of basic proportionality theorem)

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