To prove that line segment \( AC \) bisects angle \( \angle BCD \) in the parallelogram \( DABCD \) where \( AC \perp BD \), we will follow a structured approach involving properties of parallelograms, angles, and perpendicular lines.
Given:
- \( DABCD \) is a parallelogram.
- \( AC \perp BD \).
To Prove:
- \( AC \) bisects \( \angle BCD \).
Proof:
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Properties of the Parallelogram: In a parallelogram, opposite sides are equal, and opposite angles are equal. Thus, we have: \[ AD \parallel BC \quad \text{and} \quad AB \parallel CD \] Also, since \( DABCD \) is a parallelogram, we have: \[ AB = CD \quad \text{and} \quad AD = BC \]
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Focus on Points: Let the intersection of \( AC \) and \( BD \) be the point \( O \). Since \( AC \perp BD \), we have: \[ \angle AOB = \angle AOD = 90^{\circ} \]
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Show Triangles are Congruent: Since \( AC \) is a transversal across the parallel lines \( AD \) and \( BC \), the following angle relationships hold: \[ \angle ADB = \angle ABC \quad \text{(corresponding angles)} \] Therefore, we can conclude:
Triangles \( AOB \) and \( COD \) are right triangles with:
- \( AO = OC \) (since \( O \) is the midpoint because \( AC \) bisects it due to symmetry)
- \( OB = OD \) (since \( O \) is the midpoint)
Also, \( AB = CD \) (opposite sides of the parallelogram) and \( AD = BC \).
By the Hypotenuse-Leg (HL) theorem (right triangle congruence), we have: \[ \triangle AOB \cong \triangle COD \]
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Angles Relationships: Since \( \triangle AOB \cong \triangle COD \), we also get: \[ \angle OAB = \angle ODC \quad \text{and} \quad \angle OBA = \angle ODC \] This implies that: \[ \angle OBC = \angle DCO \]
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Conclusion: We have shown that \( \angle OBC = \angle DCO \), which means that line \( AC \) bisects angle \( \angle BCD \).
Thus, we conclude that: \[ AC \text{ bisects } \angle BCD. \] Hence, the proof is complete. \( \quad \blacksquare \)
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