1. **State the problem:** We need to use the parallelogram side theorem and ASA (Angle-Side-Angle) congruence to find congruent triangles in a parallelogram and show that its diagonals bisect each other. 2. **Recall the parallelogram side theorem:** Opposite sides of a parallelogram are equal in length. That is, if ABCD is a parallelogram, then $AB = DC$ and $AD = BC$. 3. **Identify triangles formed by the diagonals:** The diagonals $AC$ and $BD$ intersect at point $O$, creating four triangles: $\triangle AOB$, $\triangle COD$, $\triangle AOD$, and $\triangle BOC$. 4. **Show congruence of triangles using ASA:** Consider triangles $\triangle AOB$ and $\triangle COD$. - Side: $AB = DC$ (opposite sides of parallelogram) - 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, consider triangles $\triangle AOD$ and $\triangle BOC$:** - Side: $AD = BC$ (opposite sides of parallelogram) - 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. **Conclude that diagonals bisect each other:** Since $\triangle AOB \cong \triangle COD$, corresponding parts are equal, so $AO = CO$ and $BO = DO$. 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.