Diagonals of Rhombus Bisect Angles

Theorem
Let $OABC$ be a rhombus.

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

Proof
WLOG, we will only prove $OB$ bisects $\angle AOC$.

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

By definition of rhombus, we have:

From the above we have:

Since the angle used in the dot product is always taken to be between $0$ and $\pi$ and cosine is injective on this interval (from Shape of Cosine Function), we have:


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

The result follows.