Question

question
given: \\( \overline { c a } \\) bisects \\( \angle b a d \\) and \\( \overline { a d } \cong \overline { a b } \\).
prove: \\( \triangle a b c \cong \triangle a d c \\).

Explanation:

Step1: Definition of angle bisector

Since \(\overline{CA}\) bisects \(\angle BAD\), we have \(\angle BAC=\angle DAC\) (by the definition of an angle bisector).

Step2: Common side

\(\overline{AC}\cong\overline{AC}\) (by the reflexive property of congruence, which states that any segment is congruent to itself).

Step3: SAS (Side - Angle - Side) criterion

We are given \(\overline{AD}\cong\overline{AB}\), we found \(\angle BAC = \angle DAC\), and \(\overline{AC}\cong\overline{AC}\). So, by the SAS (Side - Angle - Side) congruence criterion, \(\triangle ABC\cong\triangle ADC\).

Answer:

\(\triangle ABC\cong\triangle ADC\) by the SAS (Side - Angle - Side) congruence criterion.