The problem requires proving that triangle $\triangle ABC$ is congruent to triangle $\triangle CDA$. Here’s the step-by-step solution:

Given

1. $AB \parallel DC$
2. $AD \parallel BC$

To Prove

$\triangle ABC \cong \triangle CDA$

Solution

1. Identify pairs of corresponding angles:

   • Since $AB \parallel DC$ and $AD \parallel BC$, the quadrilateral $ABCD$ is a parallelogram.
   • In a parallelogram, opposite angles are equal. Therefore:
     • $\angle BAC = \angle DCA$ (corresponding angles in parallel lines)
     • $\angle ADB = \angle BCA$ (corresponding angles in parallel lines)

2. Prove that diagonals bisect each other:

   • In parallelogram $ABCD$, diagonals $AC$ and $BD$ bisect each other.
   • This gives $AE = EC$ and $BE = ED$, where $E$ is the intersection of $AC$ and $BD$.

3. Use Side-Angle-Side (SAS) Congruence:

   • In $\triangle ABC$ and $\triangle CDA$:

     • Side: $AE = EC$ (since diagonals of a parallelogram bisect each other).
     • Angle: $\angle BAC = \angle DCA$ (as shown in step 1).
     • Side: $BE = ED$ (since diagonals of a parallelogram bisect each other).

   • By SAS congruence, $\triangle ABC \cong \triangle CDA$.

Conclusion

Therefore, $\triangle ABC \cong \triangle CDA$ by the Side-Angle-Side (SAS) criterion.

Would you like more details on any step or have any questions?

Related Questions

1. How do we prove that a quadrilateral is a parallelogram based on its properties?
2. Why do the diagonals of a parallelogram bisect each other?
3. Can congruent triangles always be found in any parallelogram?
4. How does the Side-Angle-Side (SAS) criterion prove congruence?
5. What other properties can be derived from a parallelogram?

Tip

When working with parallelograms, remember that opposite angles are equal, and diagonals bisect each other. This often helps in proving triangle congruence within the shape.