anonymous
  • anonymous
Use the figure below to answer the question that follows:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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anonymous
  • anonymous
What must be given to prove that ΔBHG ≅ ΔCHI? ∠GBH ≅ ∠ICH and ∠BGH ≅ ∠CIH segment BH is congruent to segment CH and segment BG is congruent to segment CI ∠GBH ≅ ∠ICH and ∠BHG ≅ ∠CHI segment BH is congruent to segment CH and segment HG is congruent to segment HI
anonymous
  • anonymous
I think it's C.

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anonymous
  • anonymous
Haha, sorry but I completely suck at math. I'm just here to say welcome back to openstudy!
anonymous
  • anonymous
It's okay, and thanks!
anonymous
  • anonymous
@e.mccormick @nincompoop @mathstudent55
anonymous
  • anonymous
I think its BH congruent to CH and BG to CL. I could be wrong because I never did this before....... ;)
anonymous
  • anonymous
Sorry if I misguide you @ilikemath50
anonymous
  • anonymous
That's okay, thanks for trying! :) @kalahkid
anonymous
  • anonymous
@e.mccormick So, what do you think?
e.mccormick
  • e.mccormick
AAA all three angles congruent only proves similarity, not congruance. |dw:1434063901941:dw|
anonymous
  • anonymous
Yeah, I know that. The same goes for SSS, ASA, SAS, and AAS right?
e.mccormick
  • e.mccormick
You always need to go with all three sides, angle-side-angle, or side-angle-side to prove congruence.
anonymous
  • anonymous
Oh, okay
anonymous
  • anonymous
I didn't know that one.
e.mccormick
  • e.mccormick
AAS... f I recall, does not always come out congruent.
e.mccormick
  • e.mccormick
There are five ways to find if two triangles are congruent: SSS, SAS, ASA, AAS and HL. http://www.mathsisfun.com/geometry/triangles-congruent-finding.html
e.mccormick
  • e.mccormick
So there is it...
anonymous
  • anonymous
Thank you, but the answer is either B or D? Right?
anonymous
  • anonymous
I'm leaning a little closer to D.
e.mccormick
  • e.mccormick
Well, they only gibe two sides, so you must use an angle out of the image that must be the same... which means that the rules for angles on crossed lines need to apply.
e.mccormick
  • e.mccormick
Note it is AAS that is congruent, not SSA.
e.mccormick
  • e.mccormick
That was the one I was trying to remember that tricks people... SSA.
e.mccormick
  • e.mccormick
Take this triangle with the marked side, side and angle: |dw:1434064547599:dw|
e.mccormick
  • e.mccormick
Now, let me modify it so those are the same, but the triangle is not: |dw:1434064605461:dw| If done carefully or with a compass, this can show that SSA/retricetriangles do not prove congruance.
anonymous
  • anonymous
Oh, okay. I'm understanding a bit better now.
e.mccormick
  • e.mccormick
|dw:1434064707031:dw|
e.mccormick
  • e.mccormick
Because the radius of the sircle determines one side, a circle can prove that retricemakes for two triangles... That was the proof I was trhying to think of.
e.mccormick
  • e.mccormick
Anyhow... one of those answers you liked was retrice not SAS.
e.mccormick
  • e.mccormick
SSA not SAS....
anonymous
  • anonymous
Oh, so it's A?
e.mccormick
  • e.mccormick
It will never be angles alone. That was the AAA problem. So you need sides. So it was one of the side ones, as you thought. So which side answer is SAS and which is SSA?
anonymous
  • anonymous
B is SAS D is SSA
e.mccormick
  • e.mccormick
segment BH is congruent to segment CH and segment BG is congruent to segment CI |dw:1434065227574:dw| segment BH is congruent to segment CH and segment HG is congruent to segment HI |dw:1434065287345:dw|
e.mccormick
  • e.mccormick
Do you see it now? I just sketched the parts involved in the answer....
anonymous
  • anonymous
Ohh, I think I'm getting it.
e.mccormick
  • e.mccormick
|dw:1434065377344:dw| |dw:1434065436305:dw|
anonymous
  • anonymous
Ugh, I have to go leave my house now. But I think I know the answer, D right?
e.mccormick
  • e.mccormick
Yah. I hope that the drawings help with remembering why it is D.
anonymous
  • anonymous
Thank you sooooo much!
e.mccormick
  • e.mccormick
Have fun!
anonymous
  • anonymous
:)

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