Dayaan M. answered • 13d
Bachelors in Computer Science with 5 years of tutoring experience
The whole proof turns on one observation: vertex A belongs to both parallelograms, and in each one the angle at A is opposite the angle you care about. So the angle at A is the bridge between ∠C and ∠G.
Check the lettering and you'll see it:
Both target angles are opposite the shared vertex. That is not a coincidence — it's how the problem was built.
The proof
| 1. ABCD and AFGH are parallelograms | Given | 2. ∠DAB ≅ ∠C | Opposite angles of a parallelogram are congruent | 3. ∠HAF ≅ ∠G | Opposite angles of a parallelogram are congruent | 4. ∠DAB ≅ ∠HAF | From the figure — see the note below | 5. ∠C ≅ ∠G | Transitive Property of Congruence (2, 3, 4) |
About step 4
Your figure isn't attached here, so I can't see exactly how the two parallelograms sit. There are two standard arrangements, and the proof works for both — only step 4's reason changes:
If AFGH shares the corner, with F on ray AB and H on ray AD, then ∠DAB and ∠HAF are literally the same angle, and step 4's reason is Reflexive Property.
If the two parallelograms sit on opposite sides of A so that the angles open away from each other, then ∠DAB and ∠HAF are vertical angles, and step 4's reason is the Vertical Angles Theorem.
Look at your diagram, pick whichever one matches, and write that reason. Steps 1, 2, 3 and 5 are identical either way.
Why transitive is the right closer
Step 5 is doing something simple that's easy to state sloppily. You have ∠C congruent to one angle at A, and ∠G congruent to an angle at A, and step 4 says those two angles at A are congruent to each other. Chain them: ∠C ≅ ∠DAB ≅ ∠HAF ≅ ∠G. Congruence is transitive, so the ends are congruent.
What you cannot do is jump straight from step 1 to step 5 by saying "both are parallelograms." Being a parallelogram tells you about angles within one figure. Nothing about ABCD alone says anything about AFGH. The shared vertex is the only thing linking them, which is why step 4 is the load-bearing step even though it looks like the throwaway one.
A good habit on proofs like this: before writing anything, find the piece the two figures have in common. It's almost always the key, and here it's vertex A.
