Diagonals of Rhombus Bisect Angles

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
, 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:

From the above we have:

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.