P and Q are the points on the sides DE and DF | vedclass

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(B) According to the Converse of the Basic Proportionality Theorem (Thales Theorem),$PQ \parallel EF$ if and only if $\frac{DP}{PE} = \frac{DQ}{QF}$.G

Vedclass · Accepted Answer

No,$PQ$ is not parallel to $EF$ because the ratios are not equal.

Vedclass · Answer

(B) According to the Converse of the Basic Proportionality Theorem (Thales Theorem),$PQ \parallel EF$ if and only if $\frac{DP}{PE} = \frac{DQ}{QF}$.Given $DP = 5 \, cm$ and $DE = 15 \, cm$,we find $PE = DE - DP = 15 - 5 = 10 \, cm$.Thus,the ratio $\frac{DP}{PE} = \frac{5}{10} = \frac{1}{2}$.Given $DQ = 6 \, cm$ and $QF = 18 \, cm$,the ratio $\frac{DQ}{QF} = \frac{6}{18} = \frac{1}{3}$.Since $\frac{DP}{PE} 
eq \frac{DQ}{QF}$ (i.e.,$\frac{1}{2} 
eq \frac{1}{3}$),the line $PQ$ is not parallel to $EF$.

$P$ and $Q$ are the points on the sides $DE$ and $DF$ of a triangle $DEF$ such that $DP = 5 \, cm$,$DE = 15 \, cm$,$DQ = 6 \, cm$,and $QF = 18 \, cm$. Is $PQ \parallel EF$? Give reasons for your answer.

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