QUESTION IMAGE
Question
given: $\angle bea \cong \angle fdg$.
prove: $\angle c$ and $\angle cbe$ are supplementary.
note: quadrilateral properties are not permitted in this proof.
step statement reason
1 $\angle bea \cong \angle fdg$ given
2 $\angle cde \cong \angle fdg$ vertical angles are congruent
note: $\overline{af}$, $\overline{cg}$ and $\overline{ac}$ are segments.
answer attempt 1 out of 2
you must answer all questions above in order to submit.
Step1: Transitive property of congruence
$\angle BEA \cong \angle CDE$
Step2: Corresponding angles imply parallel lines
$\overline{BC} \parallel \overline{BE}$ (Wait, correction: $\overline{BC} \parallel \overline{DE}$)
Wait, precise correction:
Step1: Transitive congruence
$\angle BEA \cong \angle CDE$ (from $\angle BEA \cong \angle FDG$ and $\angle CDE \cong \angle FDG$)
Step2: Corresponding angles mean parallel lines
$\overline{BC} \parallel \overline{BE}$ is wrong, correct: $\overline{BC} \parallel \overline{DE}$ because corresponding angles $\angle CDE \cong \angle BEA$
Step3: Parallel lines imply same-side interior angles supplementary
$\angle C + \angle CBE = 180^\circ$
Formal step-by-step:
Step1: Transitive congruence
$\angle BEA \cong \angle CDE$
Step2: Corresponding angles → parallel lines
$\overline{BC} \parallel \overline{DE}$
Step3: Same-side interior angles are supplementary
$\angle C + \angle CBE = 180^\circ$
Full detailed steps with proper reasons:
Step1: Apply transitive property
From $\angle BEA \cong \angle FDG$ and $\angle CDE \cong \angle FDG$, by transitive property of congruent angles, $\angle BEA \cong \angle CDE$.
Step2: Conclude parallel lines
$\angle BEA$ and $\angle CDE$ are corresponding angles. If corresponding angles are congruent, then the lines intersected by the transversal are parallel: $\overline{BC} \parallel \overline{BE}$ correction: $\overline{BC} \parallel \overline{DE}$ (transversal $\overline{AF}$ intersects $\overline{BC}$ and $\overline{DE}$)
Step3: Supplementary angles from parallel lines
When two parallel lines are cut by a transversal, same-side interior angles are supplementary. Here, $\overline{AC}$ is the transversal cutting $\overline{BC}$ and $\overline{BE}$ correction: $\overline{AC}$ cuts $\overline{BC}$ and $\overline{DE}$, but $\overline{BE}$ is on $\overline{DE}$, so $\angle C$ and $\angle CBE$ are same-side interior angles, so $\angle C + \angle CBE = 180^\circ$, meaning they are supplementary.
Formal correct steps:
Step1: Transitive congruence
$\angle BEA \cong \angle CDE$ (transitive property, since $\angle BEA \cong \angle FDG$ and $\angle CDE \cong \angle FDG$)
Step2: Corresponding angles → parallel lines
$\overline{BC} \parallel \overline{DE}$ (if corresponding angles are congruent, lines are parallel; transversal $\overline{AF}$ intersects $\overline{BC}$ and $\overline{DE}$)
Step3: Same-side interior angles supplementary
$\angle C + \angle CBE = 180^\circ$ (same-side interior angles formed by transversal $\overline{AC}$ cutting parallel lines $\overline{BC}$ and $\overline{DE}$ are supplementary)
Step3: Transitive congruence
$\angle BEA \cong \angle CDE$
Reason: Transitive Property of Congruence
Step4: Conclude parallel lines
$\overline{BC} \parallel \overline{DE}$
Reason: Corresponding Angles Converse
Step5: Supplementary angles
$\angle C + \angle CBE = 180^\circ$
Reason: Same-Side Interior Angles Theorem
Snap & solve any problem in the app
Get step-by-step solutions on Sovi AI
Photo-based solutions with guided steps
Explore more problems and detailed explanations
$\angle C$ and $\angle CBE$ are supplementary, as $\angle C + \angle CBE = 180^\circ$
Wait, to fill the proof table correctly, the missing step 3 is:
Statement: $\angle BEA \cong \angle CDE$
Reason: Transitive Property of Congruence
Then step 4:
Statement: $\overline{BC} \parallel \overline{DE}$
Reason: If corresponding angles are congruent, then lines are parallel
Then step 5:
Statement: $\angle C + \angle CBE = 180^\circ$
Reason: Same-side interior angles of parallel lines cut by a transversal are supplementary
So the immediate next step after step 2 is: