Answer:
To prove that \( AC = BD \) in the given quadrilateral \( ABCD \) where \( AB = CD \) and \(\angle ABC = \angle BCD\), we can follow these steps:
1. **Given:**
- \( ABCD \) is a quadrilateral.
- \( AB = CD \).
- \(\angle ABC = \angle BCD\).
2. **To Prove:**
- \( AC = BD \).
3. **Proof:**
- Draw diagonals \( AC \) and \( BD \) which intersect at point \( O \).
- In triangles \( \triangle ABC \) and \( \triangle DCB \):
- \( AB = CD \) (Given)
- \(\angle ABC = \angle BCD\) (Given)
- \( BC = BC \) (Common side)
- Therefore, by the SAS (Side-Angle-Side) congruence criterion, \(\triangle ABC \cong \triangle DCB\).
- Since corresponding parts of congruent triangles are equal (CPCTC), we have:
- \( AC = BD \).
Therefore, \( AC = BD \). This completes the proof.
Step-by-step explanation:
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