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$A B C D$ 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
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The correct answer is:
$1: 2$
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
$$
\begin{array}{rlrl}
\Rightarrow & & \frac{\lambda}{\lambda+1} & =\frac{1}{\mu+1} \text { and } \frac{\lambda}{\lambda+1}=\frac{\mu}{2(\mu+1)} \\
\Rightarrow & & \frac{1}{\mu+1} & =\frac{\mu}{2(\mu+1)} \\
\Rightarrow & & \frac{\lambda}{\lambda+1} & =\frac{1}{2+1}=\frac{1}{3} \\
& \therefore & 3 \lambda & =\lambda+1 \\
\Rightarrow & 2 \lambda & =1
\end{array}
$$
$$
\begin{aligned}
\Rightarrow \quad \lambda & =\frac{1}{2} \\
\text { Hence, required ratio } & =A Q: Q C=\lambda: 1 \\
& =\frac{1}{2}: 1=1: 2
\end{aligned}
$$
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