1. **State the problem:** We need to use the parallelogram side theorem and ASA (Angle-Side-Angle) congruence criterion to find congruent triangles in a parallelogram and then prove that the diagonals bisect each other. 2. **Recall the parallelogram side theorem:** Opposite sides of a parallelogram are equal in length. If the parallelogram is ABCD, then $AB = DC$ and $AD = BC$. 3. **Identify the triangles formed by the diagonals:** The diagonals $AC$ and $BD$ intersect at point $O$, creating triangles $\triangle AOB$ and $\triangle COD$, and triangles $\triangle AOD$ and $\triangle BOC$. 4. **Use ASA to prove congruence of triangles $\triangle AOB$ and $\triangle COD$:** - Side: $AB = DC$ (parallelogram side theorem) - Angle: $\angle OAB = \angle OCD$ (alternate interior angles because $AB \parallel DC$) - Angle: $\angle ABO = \angle DCO$ (alternate interior angles because $AB \parallel DC$) Thus, by ASA, $\triangle AOB \cong \triangle COD$. 5. **Similarly, prove congruence of triangles $\triangle AOD$ and $\triangle BOC$:** - Side: $AD = BC$ (parallelogram side theorem) - Angle: $\angle OAD = \angle OBC$ (alternate interior angles because $AD \parallel BC$) - Angle: $\angle ADO = \angle BCO$ (alternate interior angles because $AD \parallel BC$) By ASA, $\triangle AOD \cong \triangle BOC$. 6. **Show that diagonals bisect each other:** From the congruent triangles, corresponding parts are equal: - $AO = OC$ - $BO = OD$ Therefore, the diagonals $AC$ and $BD$ bisect each other. **Final answer:** The diagonals of a parallelogram bisect each other because the triangles formed by the diagonals are congruent by ASA using the parallelogram side theorem and alternate interior angles.