Question

Can someone check my proof please?
In the figure below, ΔABC ≅ ΔDEF. Point C is the point of intersection between segment AG and segment BF, while point E is the point of intersection between segment DG and segment BF.
http://learn.flvs.net/webdav/assessment_images/educator_geometry_v16/05_08_1b.gif
Prove ΔABC ~ ΔGEC.

Answer 1

AG intersects at point C = Given
BC ||CE = Supplementary Angles
<ACD = <GCE = Vertical Angles Theorem
<ABC = <GEC = Alternate Angles Theorem
∆ABC = ∆GEC = AA Similarity Postulate

Answer 2

I'll rewrite it in paragraph form when I know if my steps are right ^^

Answer 3

The picture seems to be unavailable

Answer 4

Hm hold on let me retry

Answer 5

Can you see the photo now? Or is still unavailable?

Answer 6

I see it, but I never learned this type of Geometry. All I can see to prove it is supplementary angles and that ACB and ECG are similar in a way. Sorry

Answer 7

Ah it's okay! Thank you for your efforts! Do you possibly know anyone who does know it?

Answer 8

Probably phi
Lots of others do, but I don't know them really

Answer 9

Alright thank you!

Answer 10

Do you recognize it? @peachpi

Answer 11

I don't think your second step is correct. BC and CE aren't parallel, they're parts of the same segment

Answer 12

Yeah I was wary of that one myself.. any ideas to what step 2 would be?

Answer 13

And for your 3rd, there is not <ACD.

I'm not sure you really need (or have enough info) to get into proportional segments. I think you can just use the angles.

1. Start with the given triangle congruency.
2 & 3. Then use the vertical angle to relate the pair at vertices C & E.
4. Use CPCTC to equate <B with <E in triangle DEF
5. Use the transitive property to equate <B with <E in triangle CEG
6. Use AA to prove similarity

Answer 14

Thank you!