$A B C D$ is a parallelogram and $P$ is the mid-point of the side $A D$. The line $B P$ meets the diagonal $A C$ in $Q$. Then, the ratio of $A Q: Q C$ is equal to

• Correct answer
  A

  $1: 2$
• B

  $2: 1$
• C

  $1: 3$
• D

  $3: 1$

Solution

Let $Q$ divides $A C$ and $B P$ in the ratio $\lambda: 1$ and $\mu: 1$ respectively. Now, point $$ Q=\frac{\lambda(b+d)+1(0)}{\lambda+1}=\frac{\lambda(b+d)}{\lambda+1}=\frac{\lambda}{\lambda+1} b+\frac{\lambda}{\lambda+1} d $$ and $\quad \mathbf{Q}=\frac{\mu \frac{\mathbf{d}}{\mathbf{2}}+\mathbf{b}}{\mu+1}=\frac{1}{\mu+1} \mathbf{b}+\frac{\mu}{2(\mu+1)} \mathbf{d}$ From Eqs. (i) and (ii), we get $$ \begin{array}{rlrl} \Rightarrow & & \frac{\lambda}{\lambda+1} & =\frac{1}{\mu+1} \text { and } \frac{\lambda}{\lambda+1}=\frac{\mu}{2(\mu+1)} \\ \Rightarrow & & \frac{1}{\mu+1} & =\frac{\mu}{2(\mu+1)} \\ \Rightarrow & & \frac{\lambda}{\lambda+1} & =\frac{1}{2+1}=\frac{1}{3} \\ & \therefore & 3 \lambda & =\lambda+1 \\ \Rightarrow & 2 \lambda & =1 \end{array} $$ $$ \begin{aligned} \Rightarrow \quad \lambda & =\frac{1}{2} \\ \text { Hence, required ratio } & =A Q: Q C=\lambda: 1 \\ & =\frac{1}{2}: 1=1: 2 \end{aligned} $$

Hence, the correct answer is option: 'A'.