Questions: Complete the proof. Statements Reasons 1. CD is the perpendicular bisector of AB 1. given 2. D is the midpoint of AB 2. 3. BD congruent to AD 3. definition of a midpoint 4. BD=AD 4. definition of congruence 5. CD=CD 5. 6. angle ADC and angle BDC are right angles 6. definition of a perpendicular bisector 7. triangle ADC and triangle BDC right triangles 7. definition of a right triangle 8. (AD)^2+(CD)^2=(AC)^2 8. Pythagorean theorem (BD)^2+(CD)^2=(BC)^2 9. substitution property of equality 9. (BD)^2+(CD)^2=(AC)^2 10. substitution property of equality 10. (BC)^2=(AC)^2 11. 11. BC=AC

Solution Steps

To complete the proof, we need to fill in the missing reasons for each statement. Here is the approach:

1. Given: $\overline{\mathrm{CD}}$ is the perpendicular bisector of $\overline{\mathrm{AB}}$.
2. D is the midpoint of $\overline{\mathrm{AB}}$ because a perpendicular bisector intersects a segment at its midpoint.
3. $\overline{\mathrm{BD}} \cong \overline{\mathrm{AD}}$ by the definition of a midpoint.
4. $B D=A D$ by the definition of congruence.
5. $C D=C D$ by the reflexive property of equality.
6. $\angle \mathrm{ADC}$ and $\angle \mathrm{BDC}$ are right angles by the definition of a perpendicular bisector.
7. $\triangle \mathrm{ADC}$ and $\triangle \mathrm{BDC}$ are right triangles by the definition of a right triangle.
8. $(A D)^{2}+(C D)^{2}=(A C)^{2}$ by the Pythagorean theorem.
9. $(B D)^{2}+(C D)^{2}=(B C)^{2}$ by the Pythagorean theorem.
10. $(B D)^{2}+(C D)^{2}=(A C)^{2}$ by the substitution property of equality.
11. $(B C)^{2}=(A C)^{2}$ by the substitution property of equality.
12. $B C=A C$ by taking the square root of both sides.

We are given that \( \overline{\mathrm{CD}} \) is the perpendicular bisector of \( \overline{\mathrm{AB}} \). This implies that point \( D \) is the midpoint of segment \( \overline{\mathrm{AB}} \).

Since \( D \) is the midpoint of \( \overline{\mathrm{AB}} \), we have: \[ \overline{\mathrm{BD}} \cong \overline{\mathrm{AD}} \] This means that the lengths of segments \( \overline{\mathrm{BD}} \) and \( \overline{\mathrm{AD}} \) are equal.

From the definition of congruence, we can express this as: \[ B D = A D \] This equality holds true because congruent segments have equal lengths.

By the reflexive property of equality, we can state: \[ C D = C D \] This is a basic property that holds for any segment.

Since \( \overline{\mathrm{CD}} \) is the perpendicular bisector, it creates right angles at points \( A \) and \( B \): \[ \angle \mathrm{ADC} \text{ and } \angle \mathrm{BDC} \text{ are right angles.} \]

This means that triangles \( \triangle \mathrm{ADC} \) and \( \triangle \mathrm{BDC} \) are right triangles.

Applying the Pythagorean theorem to both triangles, we have: \[ (A D)^2 + (C D)^2 = (A C)^2 \] \[ (B D)^2 + (C D)^2 = (B C)^2 \]

Since \( A D = B D \) and \( C D = C D \), we can substitute to find: \[ (B D)^2 + (C D)^2 = (A C)^2 \] This leads us to: \[ (B C)^2 = (A C)^2 \]

Taking the square root of both sides gives us: \[ B C = A C \]

Final Answer