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, $4CR=AB$

• A. True
• B. False

Answer 1

The correct option is A True

Given: In $△ABC$, D is mid point of BC and $DQ\parallel BA\parallel CR$

To Prove: $4CR=AB$
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))
Now, in $△ECR$ and $△EPD$

$\angle CER=\angle PED$ (Common angle)

$\angle ERC=\angle EDP$ (Corresponding angles, $CR\parallel AF$)

$\angle ECR=\angle EPD$ (Corresponding angles, $CR\parallel AF$)

Thus, $△ECR\sim △EPD$ (AAA rule)

Hence, $\frac{CE}{EP}=\frac{CR}{DP}$

$\frac{1}{2}=\frac{CR}{DP}$

Thus, $DP=2CR$...(III)
Similarly, $△CPD\sim △CAB$

$\frac{CP}{CA}=\frac{PD}{AB}$

Hence, $DP=\frac{1}{2}AB$...(IV) (Since, P is mid point of AC)
From III and IV,

$4CR=AB$