how do you solve this proof

Question

Segment EG is congruent to segment IA; Angle Ega is congruent to angle IAG

 prove that angle Gen is congruent to angle AIN

Answer 1

The key is that the two triangles share side GA, so you can get them congruent by SAS and then use CPCTC.

I'm reading the figure as quadrilateral E–G–A–I with the diagonals EA and GI crossing at N. That's what makes the angles in the "prove" line meaningful: since N sits on EA, ray EN is the same ray as EA, and since N sits on GI, ray IN is the same ray as IG. So ∠GEN is ∠GEA, and ∠AIN is ∠AIG.

The proof

1. EG ≅ IA

Given

2. ∠EGA ≅ ∠IAG

Given

3. GA ≅ AG

Reflexive Property

4. △EGA ≅ △IAG

SAS (1, 2, 3)

5. ∠GEA ≅ ∠AIG

CPCTC

6. N lies on EA and on GI

Given (from the figure)

7. ∠GEN ≅ ∠AIN

Substitution (5, 6)

Why step 4 works, and the part students lose points on

SAS only applies when the angle sits between the two sides you named. Check it:

If the given angle had been somewhere else — say ∠GEA — you would have side-side-angle, which is not a congruence shortcut, and the proof would fall apart.

Getting the correspondence right

This is where step 5 goes wrong for most people. The order of letters in step 4 is not decorative — it tells you which parts match:

△EGA ≅ △IAG  →  E ↔ I,   G ↔ A,   A ↔ G

So the angle at E corresponds to the angle at I. The angle at E in the first triangle is ∠GEA; the angle at I in the second is ∠AIG. That's exactly step 5. Write the correspondence out before you use CPCTC and you won't pair the wrong angles.

Notice the reflexive step pairs GA with AG — same segment, written in reverse order, because the vertices trade places under the correspondence. That reversal is intentional, not a typo.

If your figure places N somewhere other than the diagonal intersection, tell me where and I'll adjust the last two steps — steps 1 through 5 hold either way.