1. **Problem Statement:** Deduce that the diagonals of a rhombus meet at right angles. 2. **Definition and Properties:** A rhombus is a parallelogram with all sides equal in length. We need to prove that its diagonals intersect at 90 degrees. 3. **Approach:** Use the Side-Side-Side (SSS) congruence rule for triangles. 4. **Step 1:** Let the rhombus be ABCD with sides AB = BC = CD = DA. 5. **Step 2:** Let the diagonals AC and BD intersect at point O. 6. **Step 3:** Consider triangles AOB and COD. 7. **Step 4:** Since ABCD is a parallelogram, diagonals bisect each other, so AO = CO and BO = DO. 8. **Step 5:** Also, AB = CD (all sides equal in rhombus). 9. **Step 6:** By SSS, triangles AOB and COD are congruent because AO = CO, BO = DO, and AB = CD. 10. **Step 7:** Corresponding angles of congruent triangles are equal, so angle AOB = angle COD. 11. **Step 8:** Similarly, consider triangles AOD and BOC, which are also congruent by SSS. 12. **Step 9:** Angles AOD and BOC are equal. 13. **Step 10:** Angles AOB + BOC + COD + DOA form a full circle around point O, summing to 360 degrees. 14. **Step 11:** Since opposite angles are equal, each pair sums to 180 degrees, so angle AOB + angle BOC = 180 degrees. 15. **Step 12:** From congruence, angle AOB = angle COD and angle AOD = angle BOC, and these pairs are supplementary. 16. **Step 13:** Therefore, each angle formed by the diagonals is 90 degrees. 17. **Conclusion:** The diagonals of a rhombus intersect at right angles. $$\boxed{\text{Diagonals of a rhombus are perpendicular}}$$