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$A B C D$ is a parallelogram, with $A C, B D$ as diagonals, then $\mathbf{A C}-\mathbf{B D}$ is equal to
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Verified Answer
The correct answer is:
$2 \mathbf{A B}$
In $\triangle A B C$
$\mathbf{A C}=\mathbf{A B}+\mathbf{B C}$
and in $\triangle A B D \mathbf{A D}=\mathbf{A B}+\mathbf{B D z}$
$\mathbf{B C}=\mathbf{A B}+\mathbf{B D}$
$(\because \mathbf{A D}=\mathbf{B C}) \ldots(\mathrm{ii})$
From Eqs. (i) and (ii)
$\begin{aligned} & \mathbf{A C}=\mathbf{A B}+\mathbf{A B}+\mathbf{B D} \\ & \mathbf{A C}=\mathbf{B D}=2 \mathbf{A B}\end{aligned}$
$\mathbf{A C}=\mathbf{A B}+\mathbf{B C}$
and in $\triangle A B D \mathbf{A D}=\mathbf{A B}+\mathbf{B D z}$
$\mathbf{B C}=\mathbf{A B}+\mathbf{B D}$
$(\because \mathbf{A D}=\mathbf{B C}) \ldots(\mathrm{ii})$
From Eqs. (i) and (ii)
$\begin{aligned} & \mathbf{A C}=\mathbf{A B}+\mathbf{A B}+\mathbf{B D} \\ & \mathbf{A C}=\mathbf{B D}=2 \mathbf{A B}\end{aligned}$
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