Question

Given: start overline, A, B, end overline, \cong, start overline, C, D, end overline, comma
AB
≅
CD
, start overline, A, D, end overline, \parallel, start overline, B, C, end overline, comma
AD
∥
BC
, start overline, B, G, end overline, \cong, start overline, H, D, end overline, comma
BG
≅
HD
, angle, C, G, E, \cong, angle, A, H, F∠CGE≅∠AHF and start overline, A, E, end overline, \cong, start overline, F, C, end overline, .
AE
≅
FC
.

Prove: angle, B, A, E, \cong, angle, D, C, F∠BAE≅∠DCF.
In the proof shown in the table below, the statements and the first reason are filled in for you. Use the dropdown menu in the Reason column to select the reason for the next statement. Once you have selected a reason, a new statement will appear in a new row below. This will continue until you have selected the reason for the final statement, and then the answer area will appear with a submit button. The diagram associated with this problem is located at the bottom of the page, below the proof, and is available as a tactile printout. There may be a clarifying note below the diagram.
Step Statement Reason
1
start overline, A, B, end overline, \cong, start overline, C, D, end overline
AB
≅
CD

start overline, A, D, end overline, \parallel, start overline, B, C, end overline
AD
∥
BC

start overline, B, G, end overline, \cong, start overline, H, D, end overline
BG
≅
HD

angle, C, G, E, \cong, angle, A, H, F∠CGE≅∠AHF
start overline, A, E, end overline, \cong, start overline, F, C, end overline
AE
≅
FC

Given
2
start overline, E, F, end overline, \cong, start overline, E, F, end overline
EF
≅
EF


Reflexive Property
3
start overline, A, F, end overline, \cong, start overline, C, E, end overline
AF
≅
CE
_

Answer 1

Here's a clean proof that ∠BAE ≅ ∠DCF, written step-by-step with reasons.

Given:
- AB ≅ CD
- AD ∥ BC
- BG ≅ HD
- ∠CGE ≅ ∠AHF
- AE ≅ FC

1) EF ≅ EF. Reason: Reflexive property.

2) From AD ∥ BC, AB ≅ CD, and BG ≅ HD we get CG ≅ AH and GE ≅ HF. Reason: corresponding segment differences along the parallel lines (so the parts determined by equal whole segments and equal complementary parts are equal). (These equalities make the sides around the given equal angles match up.)

3) Triangles CGE and AHF are congruent. Reason: SAS (CG ≅ AH, GE ≅ HF, and included ∠CGE ≅ ∠AHF).

4) From congruence of triangles CGE and AHF, CE ≅ AF. Reason: CPCTC (corresponding parts of congruent triangles are congruent).

5) Now in triangles AEF and CFE we have AE ≅ FC (given), EF ≅ EF (step 1), and AF ≅ CE (step 4). So triangles AEF and CFE are congruent. Reason: SSS.

6) From congruence of triangles AEF and CFE, the angles at E that subtend corresponding sides BA and DC are equal; in particular ∠BAE ≅ ∠DCF. Reason: CPCTC (corresponding angles of congruent triangles are congruent).

Therefore ∠BAE ≅ ∠DCF. □

(Any equivalent chain of justifications that establishes CG ≅ AH and GE ≅ HF, then the congruences used above, yields the same conclusion.)