Question

Given: start overline, A, D, end overline, \cong, start overline, B, C, end overline, comma
AD
≅
BC
, start overline, A, B, end overline, \cong, start overline, C, D, end overline
AB
≅
CD
and start overline, C, E, end overline, \cong, start overline, B, E, end overline, .
CE
≅
BE
.

Prove: triangle, F, B, E, \cong, triangle, D, C, E△FBE≅△DCE.
In the table below, the first step of the proof is completed for you. To complete each following step, first use the dropdown menu in the "Statement" column to select the format of a statement. Once you select a format, a statement will appear with input boxes and/or dropdowns to complete, and a dropdown menu will appear in the reason column. Once you complete these inputs and dropdowns, navigate back to the first column and press the "try" button. If your statement and reason are valid, a new row will appear below. This process will continue until you have filled in a statement and reason for what you were asked to prove, and then the "try" button will cause the answer area to appear with a submit button. The diagram associated with this problem is located below the proof table, and is available as a tactile printout. There may be a clarifying note below the diagram.
Step Statement Reason
1
start overline, A, D, end overline, \cong, start overline, B, C, end overline
AD
≅
BC

start overline, A, B, end overline, \cong, start overline, C, D, end overline
AB
≅
CD

start overline, C, E, end overline, \cong, start overline, B, E, end overline
CE
≅
BE

Given. What are the rest of the steps, statements, and reasons for the question?

Answer 1

I will assume point F in the statement is actually point A (so △FBE means △ABE). The following completes the proof that △ABE ≅ △DCE, which gives the desired congruence.

1. AD ≅ BC; AB ≅ CD; CE ≅ BE. — Given.

2. AB ∥ CD and AD ∥ BC. — If both pairs of opposite sides of a quadrilateral are congruent, the quadrilateral is a parallelogram (so opposite sides are parallel).

3. AC and BD bisect each other (so AE ≅ CE and BE ≅ DE). — Diagonals of a parallelogram bisect each other.

4. AE ≅ DE. — From AE ≅ CE (step 3), CE ≅ BE (step 1), and BE ≅ DE (step 3); by transitivity of congruence AE = CE = BE = DE, hence AE ≅ DE.

5. AB ≅ CD; BE ≅ CE; AE ≅ DE. — Collecting the three pairwise equal sides for the two triangles.

6. △ABE ≅ △DCE. — SSS (three pairs of corresponding sides congruent).

Therefore △FBE ≅ △DCE (with F = A).