Question

given: \\( \overline { t r } \cong \overline { t s } \\), \\( \overline { a t } \\) bisects \\( \angle r t s \\). prove: \\( \triangle r u a \cong \triangle s u a \\)

Explanation:

Step1: Use the definition of angle - bisector

Since \( \overline{AT} \) bisects \( \angle RTS \), then \( \angle RTU=\angle STU \).

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

We know that \( \overline{TR}\cong\overline{TS} \) (given), \( \angle RTU = \angle STU \) (from the angle - bisector), and \( \overline{TU}\cong\overline{TU} \) (common side). So, \( \triangle RTU\cong\triangle STU \) by SAS. Then \( \overline{RU}\cong\overline{SU} \).

Step3: Consider the angles

\( \angle RUA = 180^{\circ}-\angle RUT \) and \( \angle SUA=180^{\circ}-\angle SUT \). Since \( \angle RUT=\angle SUT \) (from \( \triangle RTU\cong\triangle STU \)), then \( \angle RUA\cong\angle SUA \).

Step4: Use the SAS congruence for \( \triangle RUA \) and \( \triangle SUA \)

We have \( \overline{RU}\cong\overline{SU} \), \( \angle RUA\cong\angle SUA \), and \( \overline{UA}\cong\overline{UA} \) (common side). So, \( \triangle RUA\cong\triangle SUA \) by SAS.

Answer:

\( \triangle RUA\cong\triangle SUA \) by the SAS (Side - Angle - Side) congruence criterion.