Diagonals of Rhombus Bisect Angles/Proof 1

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 $\angle AOC$.

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 $\angle AOC$.

$\blacksquare$