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If $O$ is any point $O A+O B+O C+O D=x O E$, then find $x$, given that $A B C D$ is quadrilateral, $B$ is the point of intersection of the line joining the mid-points of opposite sides.
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Verified Answer
The correct answer is:
4
Let $P, Q, R$ and $S$ are the mid-point of sides of a quadrilateral $A B C D$ respectively.
$$
\begin{aligned}
& P=\text { mid-point of } \mathbf{A B}=\frac{\mathbf{a}+\mathbf{b}}{2} \\
& Q=\text { mid-point of } \mathbf{B C}=\frac{\mathbf{b}+\mathbf{c}}{2} \\
& R=\text { mid-point of } \mathbf{C D}=\frac{\mathbf{c}+\mathbf{d}}{2} \\
& S=\text { mid-point of } \mathbf{A D}=\frac{\mathbf{a}+\mathbf{d}}{2} \\
& \underline{\mathbf{a}+\mathbf{b}}+\frac{\mathbf{c}+\mathbf{d}}{2}
\end{aligned}
$$
Mid-point of $P R=\frac{\frac{\mathbf{a}+\mathbf{b}}{2}+\frac{\mathbf{c}+\mathbf{d}}{2}}{2}=\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}}{4}$
Mid-point of $S Q=\frac{\frac{\mathbf{a}+\mathbf{b}}{2}+\frac{\mathbf{c}+\mathbf{d}}{2}}{2}$
$$
=\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}}{4}
$$
$\therefore E=$ mid-point of $P R=$ mid-point of $S Q$
$$
\begin{aligned}
\therefore \quad \mathbf{O E}=\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}}{4} & \\
4 \mathbf{O E}=\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d} & \\
\therefore \mathbf{O A}+\mathbf{O B}+\mathbf{O C}+\mathbf{O D} & =4 \mathbf{O E} \\
x \mathbf{O E} & =4 \mathbf{O E} \\
x & =4
\end{aligned}
$$
Hence, option (1) is correct.
$$
\begin{aligned}
& P=\text { mid-point of } \mathbf{A B}=\frac{\mathbf{a}+\mathbf{b}}{2} \\
& Q=\text { mid-point of } \mathbf{B C}=\frac{\mathbf{b}+\mathbf{c}}{2} \\
& R=\text { mid-point of } \mathbf{C D}=\frac{\mathbf{c}+\mathbf{d}}{2} \\
& S=\text { mid-point of } \mathbf{A D}=\frac{\mathbf{a}+\mathbf{d}}{2} \\
& \underline{\mathbf{a}+\mathbf{b}}+\frac{\mathbf{c}+\mathbf{d}}{2}
\end{aligned}
$$
Mid-point of $P R=\frac{\frac{\mathbf{a}+\mathbf{b}}{2}+\frac{\mathbf{c}+\mathbf{d}}{2}}{2}=\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}}{4}$
Mid-point of $S Q=\frac{\frac{\mathbf{a}+\mathbf{b}}{2}+\frac{\mathbf{c}+\mathbf{d}}{2}}{2}$
$$
=\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}}{4}
$$
$\therefore E=$ mid-point of $P R=$ mid-point of $S Q$
$$
\begin{aligned}
\therefore \quad \mathbf{O E}=\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}}{4} & \\
4 \mathbf{O E}=\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d} & \\
\therefore \mathbf{O A}+\mathbf{O B}+\mathbf{O C}+\mathbf{O D} & =4 \mathbf{O E} \\
x \mathbf{O E} & =4 \mathbf{O E} \\
x & =4
\end{aligned}
$$
Hence, option (1) is correct.
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