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$A B C D$ is a parallelogram and $P$ is a point on the segment $\mathbf{A D}$ dividing it internally in the ratio $3: 1$. If the line $\mathbf{B P}$ meets the diagonal $A C$ in $Q$, then $A Q: Q C$ equals
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Verified Answer
The correct answer is:
$3: 4$
Since, $P$ divides $A D$ in the ratio $3: 1$, so $P$ will be $\frac{3 d}{4}$
Let $Q$ divides $B P$ in $1: \lambda$ and $A C$ in $1: \mu$.
$$
\begin{aligned}
& \therefore \quad \frac{\mathbf{b}+\mathbf{d}}{1+\mu}=\frac{\frac{3 \mathbf{d}}{4}+\lambda \mathbf{b}}{1+\lambda} \\
& \Rightarrow \mathbf{b}(1+\lambda)+(1+\lambda) \mathbf{d}=\frac{3}{4}(1+\mu) \mathbf{d}+\lambda(1+\mu) \mathbf{d}
\end{aligned}
$$
On equating the vectors $\mathbf{b}$ and $\mathbf{d}$, we get
$$
\begin{array}{rlrl}
& & 1+\lambda=\frac{3}{4}(1+\mu) & \text { and } 1+\lambda=\lambda(1+\mu) \\
\Rightarrow & \lambda & =\frac{3}{4} \text { and } \mu=\frac{4}{3} \\
\therefore & A Q: Q C & =1: \mu \\
& = & 1: \frac{4}{3}=3: 4
\end{array}
$$
Let $Q$ divides $B P$ in $1: \lambda$ and $A C$ in $1: \mu$.
$$
\begin{aligned}
& \therefore \quad \frac{\mathbf{b}+\mathbf{d}}{1+\mu}=\frac{\frac{3 \mathbf{d}}{4}+\lambda \mathbf{b}}{1+\lambda} \\
& \Rightarrow \mathbf{b}(1+\lambda)+(1+\lambda) \mathbf{d}=\frac{3}{4}(1+\mu) \mathbf{d}+\lambda(1+\mu) \mathbf{d}
\end{aligned}
$$
On equating the vectors $\mathbf{b}$ and $\mathbf{d}$, we get
$$
\begin{array}{rlrl}
& & 1+\lambda=\frac{3}{4}(1+\mu) & \text { and } 1+\lambda=\lambda(1+\mu) \\
\Rightarrow & \lambda & =\frac{3}{4} \text { and } \mu=\frac{4}{3} \\
\therefore & A Q: Q C & =1: \mu \\
& = & 1: \frac{4}{3}=3: 4
\end{array}
$$
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