Question

given: \\( \overline { d c } \\) is the perpendicular bisector of \\( \overline { a b } \\)
prove: \\( \overline { d a } \cong \overline { d b } \\)
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$$\begin{tabular}{|c|c|} \\hline statements & reasons \\\\ \\hline 1. \\( \\overline { d c } \\) is the perpendicular bisector of \\( \\overline { a b } \\) & 1. given \\\\ \\hline 2. \\( \\overline { a c } \\cong \\overline { c b } \\) & 2. definition of perpendicular bisector \\\\ \\hline 3. \\( \\angle d c a \\cong \\angle d c b \\) & 3. all right angles are congruent \\\\ \\hline 4. \\( \\overline { c d } \\cong \\overline { c d } \\) & 4. reflexive property \\\\ \\hline 5. & 5. \\\\ \\hline 6. & 6. \\\\ \\hline 7. \\( \\overline { d a } \\cong \\overline { d b } \\) & 7. cpctc \\\\ \\hline \\end{tabular}$$

Explanation:

Step1: Identify congruent triangles

We have two triangles \(\triangle DCA\) and \(\triangle DCB\). We know \(AC\cong CB\) (from the definition of a perpendicular bisector), \(\angle DCA\cong\angle DCB\) (all right angles are congruent), and \(CD\cong CD\) (reflexive property).

Step2: Apply the SAS (Side - Angle - Side) congruence criterion

For two triangles \(\triangle DCA\) and \(\triangle DCB\), if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent. Here, in \(\triangle DCA\) and \(\triangle DCB\), \(AC = CB\), \(\angle DCA=\angle DCB\), and \(CD = CD\). So, \(\triangle DCA\cong\triangle DCB\) by the \(SAS\) (Side - Angle - Side) congruence criterion.

Step3: Use CPCTC (Corresponding Parts of Congruent Triangles are Congruent)

Since \(\triangle DCA\cong\triangle DCB\), their corresponding parts \(DA\) and \(DB\) are congruent. That is, \(DA\cong DB\) (by \(CPCTC\)).

Answer:

1. \(\triangle DCA\cong\triangle DCB\)
2. \(SAS\)