# Prove that the diagonals of a rhombus are at right angles when O is origin c ( α ,β) B ( α + h ,β) A (h, 0)?

Aug 12, 2018

Please see the proof below

#### Explanation:

Let's work with vectors

$\vec{O A} = \left(\begin{matrix}h \\ 0\end{matrix}\right)$

$\vec{C B} = \left(\begin{matrix}h \\ 0\end{matrix}\right)$

$\vec{O C} = \left(\begin{matrix}\alpha \\ \beta\end{matrix}\right)$

$\vec{A B} = \left(\begin{matrix}\alpha \\ \beta\end{matrix}\right)$

As the figure is a rhombus

$| | \vec{O A} | | = | | \vec{C B} | | = | | \vec{O C} | | = | | \vec{A B} | |$

${h}^{2} = {\alpha}^{2} + {\beta}^{2}$

The dot product of the diagonals is

$\vec{O B} . \vec{A C} = \left(\begin{matrix}\alpha + h \\ \beta\end{matrix}\right) . \left(\begin{matrix}\alpha - h \\ \beta\end{matrix}\right)$

$= \left(\alpha + h\right) \left(\alpha - h\right) + {\beta}^{2}$

$= {\alpha}^{2} - {h}^{2} + {\beta}^{2}$

$= 0$

Therefore,

The vectors $\vec{O B}$ and $\vec{A C}$ are orthogonal, that is
the diagonals are perpendicular to one another.