# Diagonals of Rhombus Bisect Angles/Proof 2

## 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:

 $\text {(a)}: \quad$ $\ds \mathbf a + \mathbf c$ $=$ $\ds \mathbf b$ Parallelogram Law $\text {(b)}: \quad$ $\ds \norm {\mathbf a}$ $=$ $\ds \norm {\mathbf c}$

From the above we have:

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

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.

$\blacksquare$