Question

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

Explanation:

Step1: Use the Angle - Bisector Definition

Since \( \overline{BD} \) bisects \( \angle ABC \), by the definition of an angle bisector, \( \angle ABD=\angle CBD \).

Step2: Use the Given Angle

We are given \( \angle ADB = \angle CDB \).

Step3: Use the Reflexive Property

\( BD = BD \) (reflexive property of equality, a segment is congruent to itself).

Step4: Apply the ASA (Angle - Side - Angle) Congruence Criterion

In \( \triangle ABD \) and \( \triangle CBD \), we have \( \angle ABD=\angle CBD \) (from Step 1), \( BD = BD \) (from Step 3), and \( \angle ADB=\angle CDB \) (given). By the ASA (Angle - Side - Angle) congruence criterion, \( \triangle ABD\cong\triangle CBD \).

Answer:

\( \triangle ABD\cong\triangle CBD \) (by ASA congruence criterion)