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PROBLEM

Given: DECE\overline{D E} \cong \overline{C E} and FE\overline{F E} bisects DEC\angle D E C.
Prove: FAFB\overline{F A} \cong \overline{F B}.
Step
Statement
Reason
1
DECE\overline{D E} \cong \overline{C E}
FE\overline{F E} bisects DEC\angle D E C
Given
try
Type of Statement

STEP 1

1. DECE\overline{D E} \cong \overline{C E}
2. FE\overline{F E} bisects DEC\angle D E C
3. The diagram accurately represents the given information
4. Points A and B are the intersections of FD\overline{FD} and FC\overline{FC} with DC\overline{DC} respectively

STEP 2

1. Analyze the given information
2. Identify congruent triangles
3. Prove that FAFB\overline{FA} \cong \overline{FB}

STEP 3

Analyze the given information:
DECE\overline{D E} \cong \overline{C E} implies that E is the midpoint of DC\overline{DC}.
FE\overline{F E} bisects DEC\angle D E C, so DEFCEF\angle DEF \cong \angle CEF.

STEP 4

Identify congruent triangles:
We can prove that DEFCEF\triangle DEF \cong \triangle CEF using the SAS (Side-Angle-Side) congruence criterion:
1. DECE\overline{D E} \cong \overline{C E} (given)
2. DEFCEF\angle DEF \cong \angle CEF (FE\overline{F E} bisects DEC\angle D E C)
3. EF\overline{EF} is common to both triangles
Therefore, DEFCEF\triangle DEF \cong \triangle CEF

SOLUTION

Prove that FAFB\overline{FA} \cong \overline{FB}:
3.1. Since DEFCEF\triangle DEF \cong \triangle CEF, we know that FDFC\overline{FD} \cong \overline{FC}
3.2. E is the midpoint of DC\overline{DC}, so DEEC\overline{DE} \cong \overline{EC}
3.3. FDAFCB\triangle FDA \cong \triangle FCB by the SAS congruence criterion:
- FDFC\overline{FD} \cong \overline{FC} (from step 3.1)
- FDAFCB\angle FDA \cong \angle FCB (vertical angles)
- DACB\overline{DA} \cong \overline{CB} (E is midpoint of DC\overline{DC}, so DA=DE\overline{DA} = \overline{DE} and CB=CE\overline{CB} = \overline{CE})
3.4. Since FDAFCB\triangle FDA \cong \triangle FCB, we can conclude that FAFB\overline{FA} \cong \overline{FB}
Therefore, we have proved that FAFB\overline{FA} \cong \overline{FB}.

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