Diagonals of Rhombus Bisect Angles/Proof 2

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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 $\angle OCB$

RhombusBisectAngles.png


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.

$\blacksquare$