Triangle Proofs on DeltaMath - Free Printable

We are given:

> Given: $\overline{BC} \cong \overline{AD}$ and $\overline{BC} \parallel \overline{AD}$
> Prove: $\triangle ABC \cong \triangle CDA$

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## Step-by-step Proof:

We will use the Side-Angle-Side (SAS) congruence postulate.

Step 1: Use the given information

- $\overline{BC} \cong \overline{AD}$ → Given
- $\overline{BC} \parallel \overline{AD}$ → Given

These two facts tell us that in quadrilateral $ABCD$, one pair of opposite sides is both congruent and parallel. That makes $ABCD$ a parallelogram, but we don’t need to prove that — we just need to find enough to prove the triangles congruent.

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Step 2: Identify the shared side

Notice that $\overline{AC}$ is common to both $\triangle ABC$ and $\triangle CDA$. So:

> $\overline{AC} \cong \overline{CA}$ (Reflexive Property)

This gives us one pair of congruent sides.

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Step 3: Use parallel lines to find congruent angles

Since $\overline{BC} \parallel \overline{AD}$, and $\overline{AC}$ is a transversal cutting across them, then:

> $\angle BCA \cong \angle DAC$ (Alternate Interior Angles Theorem)

Why? Because when two parallel lines are cut by a transversal, alternate interior angles are congruent.

So now we have:
- Side: $\overline{BC} \cong \overline{AD}$ (given)
- Angle: $\angle BCA \cong \angle DAC$ (from parallel lines + transversal)
- Side: $\overline{AC} \cong \overline{CA}$ (reflexive)

This satisfies the SAS Congruence Postulate.

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## Final Proof Table:

| Step | Statement | Reason |
|------|------------------------------------|--------|
| 1 | $\overline{BC} \cong \overline{AD}$ | Given |
| 2 | $\overline{BC} \parallel \overline{AD}$ | Given |
| 3 | $\angle BCA \cong \angle DAC$ | Alternate Interior Angles Theorem (since $BC \parallel AD$, $AC$ transversal) |
| 4 | $\overline{AC} \cong \overline{CA}$ | Reflexive Property of Congruence |
| 5 | $\triangle ABC \cong \triangle CDA$ | SAS Congruence Postulate (Steps 1, 3, 4) |

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✔ Therefore, $\triangle ABC \cong \triangle CDA$ by SAS.

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## Bonus Insight:
This also shows that the diagonal $\overline{AC}$ divides the parallelogram into two congruent triangles — a well-known property of parallelograms.

Let me know if you’d like to see this proven using other methods (like SSS or ASA)!