In this post, we show that the diagonals of a parallelogram bisect each other. We will use the fact that the diagonal of a parallelogram is a transversal to the opposite sides of the parallelogram. The transversal and the sides form congruent alternate interior angles. Recall that the alternate interior angles formed by a transversal and parallel lines are congruent.

**Given**

Parallelogram $latex ABCD$ with diagonals $latex \overline{AC}$ and $latex \overline{BD}$.

**What to Show**

$latex \overline{AO} \cong \overline{OC}$ and $latex \overline{BO} \cong \overline{OD}$.

**Proof**

$latex \angle OAB \cong \angle OCD$ because they are alternate interior angles of parallel lines that pass through $latex \overline {AB}$ and $latex \overline {CD}$ cut by the transversal $latex \overline{AC}$ (**A**).

$latex \overline{AB} \cong \overline{CD}$ because opposite sides of a parallelogram are congruent (**S**).

$latex \angle ABO \cong \angle ODC$ because they are alternate interior angles of parallel lines that pass through $latex \overline {AB}$ and $latex \overline {CD}$ cut by the transversal $latex \overline{BD}$ (**A**).

By **ASA **Triangle Congruence theorem, $latex \triangle AOB \cong \triangle COD$.

$latex \overline{AO} \cong \overline{OC}$ and $latex \overline{BO} \cong \overline{OD}$ because corresponding sides of congruent triangles are congruent.