Diagonals of Rhombus Bisect Angles
Jump to navigation
Jump to search
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 1
Without loss of generality, we will only prove $OB$ bisects $\angle AOC$.
We have:
| \(\ds OA\) | \(=\) | \(\ds OC\) | Definition of Rhombus | |||||||||||
| \(\ds BA\) | \(=\) | \(\ds BC\) | Definition of Rhombus | |||||||||||
| \(\ds OB\) | \(=\) | \(\ds OB\) | Common Side | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \triangle OAB\) | \(\cong\) | \(\ds \triangle OCB\) | SSS |
Comparing corresponding angles gives:
- $\angle AOB = \angle COB$
hence $OB$ bisects $\angle AOC$.
$\blacksquare$
Proof 2
Without loss of generality, we will only prove $OB$ bisects $\angle AOC$.
Let the position vectors 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 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 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.
$\blacksquare$