Prove that opposite angles of a parallelogram are equal.
The correct answer and explanation is:
To prove that opposite angles of a parallelogram are equal, consider a parallelogram ABCDABCD, where AB∥CDAB \parallel CD and AD∥BCAD \parallel BC. We need to show that ∠DAB=∠BCD\angle DAB = \angle BCD and ∠ABC=∠CDA\angle ABC = \angle CDA.
Step 1: Label the Angles
Let the angles at the vertices be:
- ∠DAB=θ\angle DAB = \theta
- ∠ABC=ϕ\angle ABC = \phi
- ∠BCD=γ\angle BCD = \gamma
- ∠CDA=δ\angle CDA = \delta
Step 2: Use Parallel Line Properties
Since AB∥CDAB \parallel CD, the angle ∠DAB\angle DAB is an alternate interior angle with ∠BCD\angle BCD (because ADAD is a transversal line between the two parallel lines). By the Alternate Interior Angle Theorem, we conclude that: ∠DAB=∠BCD.\angle DAB = \angle BCD.
Therefore, θ=γ\theta = \gamma.
Step 3: Apply the Same Reasoning to the Other Pair of Opposite Angles
Similarly, since AD∥BCAD \parallel BC, the angle ∠ABC\angle ABC is an alternate interior angle with ∠CDA\angle CDA. Again, applying the Alternate Interior Angle Theorem, we get: ∠ABC=∠CDA.\angle ABC = \angle CDA.
Thus, ϕ=δ\phi = \delta.
Step 4: Conclusion
From the two observations above, we have shown that opposite angles in a parallelogram are congruent: ∠DAB=∠BCDand∠ABC=∠CDA.\angle DAB = \angle BCD \quad \text{and} \quad \angle ABC = \angle CDA.
This completes the proof.
The proof relies on the basic properties of parallel lines and transversals, specifically the Alternate Interior Angle Theorem, which ensures that alternate interior angles formed by parallel lines are equal.