1. **State the problem:** Prove that a parallelogram whose diagonals are congruent has a right angle. 2. **Given:** In parallelogram $ABCD$, $\overline{AB} \parallel \overline{CD}$ and $\overline{AD} \parallel \overline{BC}$, and the diagonals $\overline{AC} \cong \overline{DB}$. 3. **Recall theorem:** Opposite sides in a parallelogram are congruent, so $\overline{AB} \cong \overline{DC}$ and $\overline{AD} \cong \overline{BC}$. 4. **Step 1:** Since $\overline{AB} \parallel \overline{CD}$ and $\overline{AD} \parallel \overline{BC}$, by definition, $ABCD$ is a parallelogram. 5. **Step 2:** By the theorem, $\overline{AB} \cong \overline{DC}$ (opposite sides congruent). 6. **Step 3:** $\overline{AD} \cong \overline{DA}$ (reflexive property). 7. **Step 4:** Given $\overline{AC} \cong \overline{DB}$. 8. **Step 5:** By Side-Side-Side (SSS) congruence, triangles $\triangle ABC$ and $\triangle CDA$ are congruent. 9. **Step 6:** Since $\overline{AB} \parallel \overline{CD}$ and $\overline{AD} \parallel \overline{BC}$, angles $\angle 1$ and $\angle 4$ are same-side interior angles formed by transversal $\overline{AC}$. 10. **Step 7:** Same-side interior angles sum to $180^\circ$, so $m\angle 1 + m\angle 4 = 180^\circ$. 11. **Step 8:** Because triangles are congruent, $m\angle 1 = m\angle 4$, so $2 \cdot m\angle 1 = 180^\circ$. 12. **Step 9:** Divide both sides by 2 to find $m\angle 1 = 90^\circ$. **Final answer:** The parallelogram has a right angle, $m\angle 1 = 90^\circ$.