Question

State true or false:

The side $AC$ of a triangle $ABC$ is produced to point $E$ so that $CE=\frac{1}{2}AC$. $D$ is the mid-point of $BC$ and $ED$ produced meets $AB$ at $F$. Lines through $D$ and $C$ are drawn parallel to $AB$ which meet $AC$ at point $P$ and $EF$ at point $R$ respectively. Hence, $3DF=EF$

• A. True
• B. False

Answer 1

The correct option is A True

Given: In $△ABC$, D is mid point of BC and $DP\parallel BA\parallel CR$
To Prove: 3 DF = EF
In $△ABC$, D is the midpoint of BC and DP are drawn parallel to BA.
Therefore, P is the midpoint of AC.
$AP=PC$

Now, $FA\parallel DP\parallel RC$ and APC is transversal such that AP = PC and FDR is the another transversal
Hence, $FD=DR$ .........(I) (by intercept theorem)
$EC=\frac{1}{2}AC=PC$
In $△EPD$,
C is the midpoint of EP and $CR\parallel DP$ .
R must be the midpoint of DE.
Thus, $DR=RE$ .....(II)
Hence, $FD=DR=RE$ (from (I) and (II))
$FE=3DF$