Diagonals of Rhombus Bisect Angles

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Theorem

Let $OABC$ be a rhombus.

Then:

$(1): \quad OB$ bisects $\angle AOC$ and $\angle ABC$
$(2): \quad AC$ bisects $\angle OAB$ and $\angle OCB$


Proof

Without loss of generality, we will only prove $OB$ bisects $\angle AOC$.


Let the position vector of $A$, $B$ and $C$ with respect to $O$ be $\mathbf a$, $\mathbf b$ and $\mathbf c$ respectively.

By definition of rhombus, we have:

\((a):\quad\) \(\displaystyle \mathbf a + \mathbf c\) \(=\) \(\displaystyle \mathbf b\) $\quad$ Parallelogram Law $\quad$
\((b):\quad\) \(\displaystyle \left \vert \mathbf a \right \vert\) \(=\) \(\displaystyle \left \vert \mathbf c \right \vert\) $\quad$ $\quad$

From the above we have:

\(\displaystyle \cos \angle \mathbf a, \mathbf b\) \(=\) \(\displaystyle \frac {\mathbf a \cdot \mathbf b} {\left \vert \mathbf a \right \vert \left \vert \mathbf b \right \vert}\) $\quad$ Definition 2 of Dot Product $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\mathbf a \cdot \left({\mathbf a + \mathbf c}\right)} {\left \vert \mathbf a \right \vert \left \vert \mathbf b \right \vert}\) $\quad$ from $(a)$ above: $\mathbf b = \mathbf a + \mathbf c$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\mathbf a \cdot \mathbf a + \mathbf a \cdot \mathbf c} {\left \vert \mathbf a \right \vert \left \vert \mathbf b \right \vert}\) $\quad$ Dot Product Distributes over Addition $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac { {\left \vert \mathbf a \right \vert}^2 + \mathbf a \cdot \mathbf c} {\left \vert \mathbf a \right \vert \left \vert \mathbf b \right \vert}\) $\quad$ Dot Product of Vector with Itself $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac { {\left \vert \mathbf c \right \vert}^2 + \mathbf a \cdot \mathbf c} {\left \vert \mathbf c \right \vert \left \vert \mathbf b \right \vert}\) $\quad$ from $(b)$ above: $\left \vert \mathbf a \right \vert = \left \vert \mathbf c \right \vert$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\mathbf c \cdot \mathbf c + \mathbf a \cdot \mathbf c} {\left \vert \mathbf c \right \vert \left \vert \mathbf b \right \vert}\) $\quad$ Dot Product of Vector with Itself $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\mathbf c \cdot \left({\mathbf a + \mathbf c}\right)} {\left \vert \mathbf c \right \vert \left \vert \mathbf b \right \vert}\) $\quad$ Dot Product Distributes over Addition $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\mathbf c \cdot \mathbf b} {\left \vert \mathbf c \right \vert \left \vert \mathbf b \right \vert}\) $\quad$ from $(a)$ above: $\mathbf b = \mathbf a + \mathbf c$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \cos \angle \mathbf c, \mathbf b\) $\quad$ Definition 2 of Dot Product $\quad$

By definition of dot product, the angle between the vectors is between $0$ and $\pi$.

From Shape of Cosine Function, cosine is injective on this interval.

Hence:

$\angle \mathbf a, \mathbf b = \angle \mathbf c, \mathbf b$

The result follows.

$\blacksquare$