Question

4CR = AB
If the above statement is true then mention answer as 1, else mention 0 if false

Answer 1

Given: In $△ABC$, D is mid point of BC and $DQ\parallel BA\parallel CR$
To Prove: 4 CR = 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, $CR=2DP$...(III)
Similarly, $△CPD\sim △CAB$
$\frac{CP}{CA}=\frac{PD}{AB}$
Hence, $DP=2AB$...(IV) (Since, P is mid point of AC)
From III and IV,
$CR=4AB$