Question

asa proof #2 given: $overline{ac}$ bisects $angle bad$, $overline{ac}$ bisects $angle bcd$ prove: $\triangle baccong\triangle dac$

Explanation:

Step1: Define angle - bisector property

Since $\overline{AC}$ bisects $\angle BAD$, we have $\angle BAC=\angle DAC$ (by the definition of an angle - bisector, which divides an angle into two equal angles).

Step2: Another angle - bisector property

Since $\overline{AC}$ bisects $\angle BCD$, we have $\angle BCA=\angle DCA$ (by the definition of an angle - bisector).

Step3: Common side

$\overline{AC}=\overline{AC}$ (by the reflexive property of equality, which states that any segment is equal to itself).

Step4: Apply ASA congruence

In $\triangle BAC$ and $\triangle DAC$, we have $\angle BAC=\angle DAC$, $\overline{AC}=\overline{AC}$, and $\angle BCA=\angle DCA$. So, $\triangle BAC\cong\triangle DAC$ by the Angle - Side - Angle (ASA) congruence criterion.

Statements	Reasons
$\angle BAC=\angle DAC$	Definition of angle - bisector
$\overline{AC}$ bisects $\angle BCD$	Given
$\angle BCA=\angle DCA$	Definition of angle - bisector
$\overline{AC}=\overline{AC}$	Reflexive property of equality
$\triangle BAC\cong\triangle DAC$	ASA congruence criterion

Answer:

The proof is completed as shown in the table above, demonstrating that $\triangle BAC\cong\triangle DAC$ by ASA.