Sure! Here's the completed two-column proof with the statements and reasons appropriately matched.
Proof that parallelogram ABCD with congruent diagonals DB and AC is a rectangle:
| Statement | Reason | |---------------------------------------------------------------------------|------------------------------------------------------------| | 1. ABCD is a parallelogram. | Given | | 2. DB ≅ AC | Given | | 3. DA ≅ CB | Opposite sides of a parallelogram are congruent. | | 4. AB ≅ AB | Reflexive Property of Congruence | | 5. ΔABD ≅ ΔACB | SSS Congruence Theorem | | 6. ∠BAD ≅ ∠ABC | CPCTC Theorem of Congruence | | 7. m∠BAD = m∠ABC | Definition of congruence | | 8. m∠BAD + m∠ABC = 180° | Consecutive angles of a parallelogram are supplementary | | 9. m∠BAD + m∠BAD = 180° | Substitution Property of Equality | | 10. 2m∠BAD = 180° | Combine like terms | | 11. m∠BAD = 90° | Division Property of Equality | | 12. m∠ABC = 90° | Transitive Property of Equality | | 13. m∠ADC = m∠DAB = 90° | Opposite angles of a parallelogram are congruent | | 14. ∠BAD, ∠DCB, ∠ABC, and ∠ADC are right angles | Definition of a right angle | | 15. ABCD is a rectangle | Definition of a rectangle |
This arrangement clearly lays out the logical flow from the properties of parallelograms to the conclusion that ABCD is a rectangle.