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Questions  >>  Olympiad-Math  >>  Class 10
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Q)

In the given figure if $BP \parallel CF,DP \parallel EF,$ then $AD : DE$ is equal to

$\begin{array}{1 1} 3:4 \\ 1:3 \\ 1:4 \\ 2:3 \end{array} $

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A)
Solution :
Since $BD \parallel CF$
Then , $\large\frac{AP}{PF}=\frac{AB}{BC}$
[Using Thales Theorem]
=> Again , Since $DP \parallel DF$
Then , $\large\frac{AP}{PF} =\frac {AD}{DE}$
Then $\large\frac{AD}{DE} =\frac{1}{3}$
$AD:DE =1:3$
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