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.
​

Answer 1

${\huge{\underline{\underline{\rm{QuestiOn:}}}}}$

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.

${\huge{\underline{\underline{\rm{SolutiOn:}}}}}$

${\bf{Proof:}}$

In ∆XDE , PQ || DE .........(Given)

∴

${ \huge{ \frac{XP}{XQ} = \frac{PD}{DE} }}..............(1)$

(Basic proportionally Theorem)

In ∆XDE , QR || EF

∴

${ \huge{ \frac{XR}{RF} = \frac{XQ}{QE} }}.............(2)$

(Basic proportionally Theorem)

∴From equation(1)&(2)

${ \huge{ \frac{XP}{PD} = \frac{XR}{RF} }}$

∴seg PR|| seg DE......

(converse of basic proportionally Theorem)

_____________________Proved...

Answer 2

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