Question

multiple choice 4 points
what other information do you need in order to prove the triangles congruent using sas?
triangle image with vertices a, b, c, d
options:

• $overline{ac} cong overline{bd}$
• $angle bac cong angle dac$
• $overline{ac} perp overline{bd}$
• $angle cba cong angle cda$

5 multiple choice 5 points
what else must you know to prove the triangles congruent by sas?
quadrilateral/triangle image with vertices a, b, c, d
options:

• $overline{ad} cong overline{ac}$
• $ab cong bc$
• $ab cong cd$
• $overline{ad} cong overline{bc}$

Explanation:

Question 4 (Triangles Congruent via SAS)

Step1: Recall SAS Congruence

SAS (Side - Angle - Side) congruence requires two sides and the included angle of one triangle to be congruent to two sides and the included angle of another triangle. In the given triangle diagram (with \( \triangle ABC \) and \( \triangle ADC \), sharing \( AC \), and \( BC = CD \) as marked), we need the included angle between the known side (\( AC \)) and the equal sides (\( BC, CD \)) to be congruent.

Step2: Analyze Options

• Option \( \overline{AC}\cong\overline{BD} \): Not relevant to SAS for these triangles.
• Option \( \angle BAC\cong\angle DAC \): \( AC \) is common, \( BC = CD \), and \( \angle BAC \) and \( \angle DAC \) are the included angles. This satisfies SAS.
• Option \( \overline{AC}\perp\overline{BD} \): Perpendicularity is not needed for SAS.
• Option \( \angle CBA\cong\angle CDA \): This is not the included angle for SAS.

Step1: Recall SAS Congruence

SAS needs two sides and the included angle. In the parallelogram - like diagram (assuming \( ABCD \) is a parallelogram - like figure with \( \angle DAC \) and \( \angle BCA \) marked as equal), we need the sides adjacent to the equal angle to be equal.

Step2: Analyze Options

• Option \( \overline{AD}\cong\overline{AC} \): Not relevant to the triangles formed by \( AC \).
• Option \( \overline{AB}\cong\overline{BC} \): Not related to the triangles with the equal angle.
• Option \( \overline{AB}\cong\overline{CD} \): In a parallelogram \( AB = CD \), but for SAS of the triangles (e.g., \( \triangle ADC \) and \( \triangle BCA \)), we need \( AD = BC \) (since \( \angle DAC=\angle BCA \), and \( AC \) is common). Wait, looking at the options again, if we consider the triangles with the marked angles, the sides adjacent to the equal angle: if \( AD = BC \), \( \angle DAC=\angle BCA \), and \( AC \) is common, then SAS holds. Among the options, \( \overline{AD}\cong\overline{BC} \) (the last option, probably a typo in the original, but assuming the last option is \( \overline{AD}\cong\overline{BC} \)) or if we re - check, in the given options, the correct one for SAS (with the marked angles) is \( \overline{AD}\cong\overline{BC} \) (the last option as per the layout, assuming the options are \( \overline{AD}\cong\overline{AC} \), \( \overline{AB}\cong\overline{BC} \), \( \overline{AB}\cong\overline{CD} \), \( \overline{AD}\cong\overline{BC} \)). Since in the triangles formed by \( AC \), \( \angle DAC=\angle BCA \), \( AC \) is common, and we need \( AD = BC \) for SAS.

Answer:

\( \boldsymbol{\angle BAC\cong\angle DAC} \)

Question 5 (Triangles Congruent via SAS)