it is given that $\overline{ab} \cong \overline{dc}$ and $\angle abc \cong \angle dcb$. $\overline{bc}$ is congruent to itself by the $\boxed{}$ property and in both $\triangle abc$ and $\boxed{}$. two sides of $\triangle abc$ and their non-included angle are congruent to two sides of $\triangle dbc$ and their $\boxed{}$ angle, so $\boxed{}$ by the $\boxed{}$ triangle congruency theorem. corresponding parts of congruent triangles are congruent, so $\boxed{}$. from the figure, $\boxed{}$ because they are vertical angles. two angles and $\boxed{}$ side of $\triangle aeb$ are congruent to two angles and the corresponding $\boxed{}$ side of $\triangle dec$ so $\boxed{}$ by the $\boxed{}$ triangle congruency theorem. corresponding parts of congruent triangles are congruent, so $\overline{ae} \cong \overline{de}$.

Question

Accepted Answer

# Explanation:
## Step1: Identify the Reflexive Property
The segment $ \overline{BC} $ is congruent to itself by the **Reflexive Property** (a segment is congruent to itself).

## Step2: Identify the Second Triangle
We are comparing $ \triangle ABC $ and $ \triangle DCB $ (since $ \overline{AB} \cong \overline{DC} $, $ \angle ABC \cong \angle DCB $, and $ \overline{BC} \cong \overline{CB} $).

## Step3: Identify the Angle Type
The angle between $ \overline{AB} $ and $ \overline{BC} $ in $ \triangle ABC $, and between $ \overline{DC} $ and $ \overline{CB} $ in $ \triangle DCB $ is the **included** angle (wait, the given option was "non - included", but actually, for SAS, it's included. But following the problem's given "non - included" might be a typo, but we proceed with the problem's context). Wait, no, in $ \triangle ABC $, sides $ AB $ and $ BC $, angle $ \angle ABC $; in $ \triangle DCB $, sides $ DC $ and $ CB $, angle $ \angle DCB $. So the angle is included between the two sides. But the problem has "non - included" selected. Maybe a mistake, but we follow the problem's structure.

## Step4: Apply SAS Congruence
Two sides and the included (or as per problem, non - included, but actually SAS is included) angle: $ \overline{AB} \cong \overline{DC} $, $ \angle ABC \cong \angle DCB $, $ \overline{BC} \cong \overline{CB} $. So by **SAS (Side - Angle - Side)** Congruence Theorem, $ \triangle ABC \cong \triangle DCB $.

## Step5: Corresponding Parts (CPCTC)
By CPCTC, $ \overline{AC} \cong \overline{DB} $ (corresponding sides of congruent triangles).

## Step6: Vertical Angles
$ \angle AEB \cong \angle DEC $ because they are vertical angles.

## Step7: AAS or ASA Congruence for $ \triangle AEB $ and $ \triangle DEC $
We have $ \angle AEB \cong \angle DEC $, $ \angle EAB \cong \angle EDC $ (from $ \triangle ABC \cong \triangle DCB $, so corresponding angles), and $ \overline{AB} \cong \overline{DC} $ (or $ \overline{AE} $ and $ \overline{DE} $ related). Wait, more accurately, from $ \triangle ABC \cong \triangle DCB $, we have $ \angle BAC \cong \angle CDB $, $ \overline{AC} \cong \overline{DB} $. Then $ \angle AEB \cong \angle DEC $ (vertical angles), and $ \angle EAB \cong \angle EDC $, $ \overline{AB} \cong \overline{DC} $. So by **AAS (Angle - Angle - Side)** or **ASA (Angle - Side - Angle)**. Let's see: $ \angle AEB \cong \angle DEC $ (vertical), $ \angle EAB \cong \angle EDC $ (from $ \triangle ABC \cong \triangle DCB $), and $ \overline{AB} \cong \overline{DC} $. So AAS. Then $ \triangle AEB \cong \triangle DEC $ by AAS.

## Step8: Final CPCTC for $ \overline{AE} \cong \overline{DE} $
Since $ \triangle AEB \cong \triangle DEC $, by CPCTC, $ \overline{AE} \cong \overline{DE} $.

Filling in the blanks:
1. $ \overline{BC} $ is congruent to itself by the **Reflexive** Property.
2. In both $ \triangle ABC $ and $ \triangle DCB $.
3. Two sides of $ \triangle ABC $ and their non - included (should be included, but as per problem) angle are congruent to two sides of $ \triangle DBC $ (wait, $ \triangle DCB $) and their non - included angle.
4. So $ \triangle ABC \cong \triangle DCB $ by the **SAS** Triangle Congruency Theorem.
5. Corresponding parts of congruent triangles are congruent, so $ \overline{AC} \cong \overline{DB} $.
6. From the figure, $ \angle AEB \cong \angle DEC $ because they are vertical angles.
7. Two angles and a side of $ \triangle AEB $ are congruent to two angles and the corresponding side of $ \triangle DEC $ so $ \triangle AEB \cong \triangle DEC $ by the **AAS** (or ASA) Triangle Congruency Theorem.

# Answer:
1. Reflexive
2. $ \triangle DCB $
3. (as per problem, non - included, but correct is included)
4. $ \triangle ABC \cong \triangle DCB $
5. SAS
6. $ \overline{AC} \cong \overline{DB} $
7. $ \angle AEB \cong \angle DEC $
8. (a)
9. (corresponding)
10. $ \triangle AEB \cong \triangle DEC $
11. AAS (or ASA)

(Note: There were some possible typos in the problem, but this is the step - by - step breakdown based on triangle congruence concepts.)

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QUESTION IMAGE

Question

Explanation:

Step1: Identify the Reflexive Property

Step2: Identify the Second Triangle

Step3: Identify the Angle Type

Step4: Apply SAS Congruence

Step5: Corresponding Parts (CPCTC)

Step6: Vertical Angles

Step7: AAS or ASA Congruence for \( \triangle AEB \) and \( \triangle DEC \)

Step8: Final CPCTC for \( \overline{AE} \cong \overline{DE} \)

Answer: