Question

given: \\( \overline { d c } \\) is the perpendicular bisector of \\( \overline { a b } \\)
prove: \\( \overline { d a } \cong \overline { d b } \\)

statements	reasons
2. \\( \overline { a c } \cong \overline { c b } \\)	2.
3. \\( \angle d c a \\) and \\( \angle d c b \\) are right angles	3. definition of perpendicular bisector
4. \\( \angle d c a \cong \angle d c b \\)	4. all right angles are congruent
5. \\( \overline { c d } \cong \overline { c d } \\)	5. reflexive property
6.	6. sas
7. \\( \overline { d a } \cong \overline { d b } \\)	7. cpctc

Explanation:

Step1: Reason for \( \overline{AC}\cong\overline{CB} \)

Since \( DC \) is the perpendicular bisector of \( \overline{AB} \), by the definition of a mid - point (a point on a line segment that divides it into two equal parts), if \( C \) is on the perpendicular bisector of \( \overline{AB} \), then \( C \) is the mid - point of \( \overline{AB} \). So the reason for \( \overline{AC}\cong\overline{CB} \) is "Definition of Midpoint".

Step2: Triangle Congruence

We have \( \overline{AC}\cong\overline{CB} \) (from step 1), \( \angle DCA\cong\angle DCB \) (all right angles are congruent), and \( \overline{CD}\cong\overline{CD} \) (reflexive property). By the Side - Angle - Side (SAS) congruence criterion (\( SAS \): if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the two triangles are congruent), the triangles \( \triangle DCA\) and \( \triangle DCB \) are congruent. So the statement for step 6 is \( \triangle DCA\cong\triangle DCB \).

Answer:

1. Definition of Midpoint; 6. \( \triangle DCA\cong\triangle DCB \)