given: $ \overline { b d } $ bisects $ \angle a d c $
$ \overline { b d } $ bisects $ \angle a b c $
prove: $ \triangle d a b \cong \triangle d c b $

Question

Accepted Answer

# Explanation:
## Step1: Definition of angle bisector
Since $ \overline{BD} $ bisects $ \angle ADC $, $ \angle ADB=\angle CDB $.
Since $ \overline{BD} $ bisects $ \angle ABC $, $ \angle ABD = \angle CBD $.
## Step2: Common side
$ BD = BD $ (reflexive property of equality).
## Step3: ASA (Angle - Side - Angle) congruence criterion
In $ \triangle DAB $ and $ \triangle DCB $, we have $ \angle ADB=\angle CDB $, $ BD = BD $, $ \angle ABD=\angle CBD $. By the ASA (Angle - Side - Angle) congruence criterion, $ \triangle DAB\cong\triangle DCB $.

# Answer:
| Statements | Reasons |
| --- | --- |
| $ \overline{BD} $ bisects $ \angle ADC $, $ \overline{BD} $ bisects $ \angle ABC $ | Given |
| $ \angle ADB=\angle CDB $, $ \angle ABD=\angle CBD $ | Definition of angle bisector |
| $ BD = BD $ | Reflexive property of equality |
| $ \triangle DAB\cong\triangle DCB $ | ASA (Angle - Side - Angle) congruence criterion |

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QUESTION IMAGE

Question

Explanation:

Step1: Definition of angle bisector

Step2: Common side

Step3: ASA (Angle - Side - Angle) congruence criterion

Answer:

Statements	Reasons
\( \angle ADB=\angle CDB \), \( \angle ABD=\angle CBD \)	Definition of angle bisector
\( BD = BD \)	Reflexive property of equality
\( \triangle DAB\cong\triangle DCB \)	ASA (Angle - Side - Angle) congruence criterion