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1990 Paper 1 Q6
D: 1500.0 B: 1505.5
vectors parallelogram geometric proof dot product projection area diagonal properties

Let \(ABCD\) be a parallelogram. By using vectors, or otherwise, prove that:

  1. \(AB^{2}+BC^{2}+CD^{2}+DA^{2}=AC^{2}+BD^{2}\);
  2. \(|AC^{2}-BD^{2}|\) is 4 times the area of the rectangle whose sides are any side of the parallelogram and the projection of an adjacent side on that side.
State and prove a result like \((ii)\) about \(|AB^{2}-AD^{2}|\) and the diagonals.

Solution: Set up coordinates such that \(A\) at the origin and \(\vec{AB} = \mathbf{x}\) and \(\vec{AD} = \mathbf{y}\) and so \(\vec{AC} = \mathbf{x}+\mathbf{y}\)

  1. \begin{align*} AC^2 + BD^2 &= (\mathbf{x}+\mathbf{y})\cdot (\mathbf{x}+\mathbf{y}) + (\mathbf{y}-\mathbf{x})\cdot(\mathbf{y}-\mathbf{x}) \\ &= 2\mathbf{x}\cdot\mathbf{x} + 2\mathbf{y}\cdot\mathbf{y} \\ &= AB^2 + CD^2 +AD^2 + BC^2 \end{align*}
  2. \begin{align*} AC^2 -BD^2 &= (\mathbf{x}+\mathbf{y})\cdot (\mathbf{x}+\mathbf{y}) - (\mathbf{y}-\mathbf{x})\cdot(\mathbf{y}-\mathbf{x}) \\ &= 4 \mathbf{x}\cdot \mathbf{y} \end{align*} \(\mathbf{x}\cdot\mathbf{y} = |\mathbf{x}||\mathbf{y}|\cos \theta\) which is exactly the lenth of one side mutliplied by the length of the projection to that same side.
\begin{align*} AB^2 - AD^2 &= \mathbf{x}\cdot\mathbf{x} - \mathbf{y}\cdot \mathbf{y} \\ &= (\mathbf{x}+\mathbf{y})\cdot(\mathbf{x}-\mathbf{y}) \\ &= AC \cdot BD \end{align*} So this is the area of the rectangle formed by the length of one diagonal and the projection of the other diagonal onto it.

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