Question

Given: A, B, C, DABCD is a parallelogram and start overline, A, C, end overline, is perpendicular to, start overline, B, D, end overline, .
AC
⊥
BD
.

Prove: start overline, A, C, end overline
AC
bisects angle, B, C, D∠BCD.

Answer 1

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:

1. **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:
\[
AB = CD \quad \text{and} \quad AD = BC
\]

2. **Focus on Points**:
Let the intersection of \( AC \) and \( BD \) be the point \( O \). Since \( AC \perp BD \), we have:
\[
\angle AOB = \angle AOD = 90^{\circ}
\]

3. **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:
\[
\triangle AOB \cong \triangle COD
\]

4. **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:
\[
\angle OBC = \angle DCO
\]

5. **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. \( \quad \blacksquare \)