1. **State the problem:** Given parallelogram ABCD with diagonals AC and BD perpendicular ($AC \perp BD$), prove that $AB \cong BC$. 2. **Recall properties:** In a parallelogram, diagonals bisect each other. Also, if diagonals are perpendicular, triangles formed are right triangles. 3. **Use the given:** $ABCD$ is a parallelogram and $AC \perp BD$. 4. **Diagonals bisect each other:** Let $E$ be the intersection of diagonals $AC$ and $BD$. Then $AE = CE$ and $BE = DE$. 5. **Triangles to consider:** Consider triangles $ABE$ and $CBE$. 6. **Show congruence:** - $AE = CE$ (diagonal bisector) - $BE = BE$ (reflexive property) - $\angle AEB = \angle CEB = 90^\circ$ (given $AC \perp BD$) 7. **Apply RHS (Right angle-Hypotenuse-Side) congruence:** Triangles $ABE$ and $CBE$ are congruent. 8. **Conclude:** Corresponding sides $AB$ and $BC$ are congruent, so $AB \cong BC$. **Final answer:** $AB \cong BC$