Question

3 directions - complete the proof below (answers may be used more than once):
given:
$overline{ac}$ bisects $angle bad$.
$angle bcongangle d$
prove: $\triangle abccong\triangle adc$
$overline{ac}$ bisects $angle bad$ def. of midpoint
$angle baccongangle dac$ def. of bisector
$overline{ac}congoverline{ac}$ reflexive property
$angle bcongangle d$ given
$\triangle abccong\triangle adc$ aas

Explanation:

Step1: Use the definition of an angle bisector

Since \( \overline{AC} \) bisects \( \angle BAD \), by the definition of an angle bisector, \( \angle BAC\cong\angle DAC \).

Step2: Use the reflexive property

For any segment \( \overline{AC} \), \( \overline{AC}\cong\overline{AC} \) (reflexive property of congruence).

Step3: Use the AAS (Angle - Angle - Side) congruence criterion

We are given \( \angle B\cong\angle D \), \( \angle BAC\cong\angle DAC \), and \( \overline{AC}\cong\overline{AC} \). In the AAS congruence criterion, if two angles and a non - included side of one triangle are congruent to two angles and the corresponding non - included side of another triangle, then the two triangles are congruent. Here, in \( \triangle ABC \) and \( \triangle ADC \), \( \angle B\) and \( \angle D\) are angles, \( \angle BAC\) and \( \angle DAC\) are angles, and \( \overline{AC}\) is the non - included side.

Answer:

The proof is completed as follows:

• \( \overline{AC} \) bisects \( \angle BAD\) is given.
• \( \angle BAC\cong\angle DAC\) (Def. of Bisector).
• \( \overline{AC}\cong\overline{AC}\) (Reflexive Property).
• \( \angle B\cong\angle D\) (Given).
• \( \triangle ABC\cong\triangle ADC\) (AAS).