anonymous
  • anonymous
ABCD is a square, M the half point of AB. With center on AB an arc is described that cuts AD in Q and BC in P. Prove that triangle MBP = triangle MAQ
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
|dw:1333162114353:dw|
anonymous
  • anonymous
|dw:1333162179190:dw|
anonymous
  • anonymous
this will help u a lot!!

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anonymous
  • anonymous
Thank you.
experimentX
  • experimentX
prove those two triangles are congruent 1) both 90 degree 2) midpoint bisects so bases are equal 3) radii of circle are equal ... equal hypotenuse
anonymous
  • anonymous
I think I need to prove another theorem first, But it's totally right what you say experimentX, thankyou!
Directrix
  • Directrix
1. ABCD is a square 2.
Directrix
  • Directrix
@No-data --> I think I need to add a statement in the proof to establish that an arc was drawn. Otherwise, there is no basis for the radii of two circles statement seemingly coming "out of nowhere." Let me know your thoughts after you have time to think about the proof. Thanks.
anonymous
  • anonymous
@Rohangrr WOOOAAA that's a long answer!
anonymous
  • anonymous
i type really fast
anonymous
  • anonymous
You're totally right Directrix!. There is no need for that statement, because it's given in the problem. But I did not write that part sorry =P
anonymous
  • anonymous
The book I copied it from is in spanish and I was too lazy to translate everything.
Directrix
  • Directrix
@No-data --> Given information should be part of a proof and introduced into the proof where it makes for the best logical fit. At least, that's what I was taught. Does the proof make sense to you? Isn't it interesting that the HL Theorem for proving right triangles congruent describes a scenario in which two sides and the none-included angle of a triangle determine a unique triangle. If the triangle is not right, then SSA becomes the basis for the "ambiguous case" in trigonometry.
anonymous
  • anonymous
I have a mental block with proofs.
anonymous
  • anonymous
The proof makes sense. But I'm no quite sure about the ambiguous case you're talking about.
anonymous
  • anonymous
Actually the scenario is two sides and the angle between them, I mean hypo and any of the other sides.
Directrix
  • Directrix
|dw:1333172158917:dw|
anonymous
  • anonymous
I'm sorry Directrix I dont' understand what you're trying to say. What I'm sure about is that two sides of a triangle an the angle between them it's enough to determine it.
anonymous
  • anonymous
and for the the HL theorem these sides are the hipo, other side, and the right angle.
Directrix
  • Directrix
Angles A and M are the congruent angle pair and not not included between the two pairs of congruent sides QM, AM and then MP, BM. The ambiguous case occurs when one uses the law of sines to determine missing measures of a triangle when given two sides and an angle opposite one of those angles (SSA). In this ambiguous case, three possible situations can occur: 1) no triangle with the given information exists, 2) one such triangle exists, or 3) two distinct triangles may be formed that satisfy the given conditions. http://jwilson.coe.uga.edu/emt668/emat6680.2001/mealor/emat%206700/law%20of%20sines/law%20of%20sines%20ambiguous%20case/lawofsinesambiguouscase.html
Directrix
  • Directrix
--> What I'm sure about is that two sides of a triangle an the angle between them it's enough to determine it. Reply: I agree - that is the SAS Triangle Congruence Theorem. That cannot be used in this proof because the angles we have congruent are not the included angles between the pairs of congruent sides.
anonymous
  • anonymous
Yes you're right Directrix. I think my brain is not totally functional right know, I'll better go to bed. But I'll think about it an give you an answer. I promise. Thank you for the help!.
Directrix
  • Directrix
I look forward to the progression of the discussion. Thanks.
anonymous
  • anonymous
I still don't get what you wanted to say @Directrix but I already understood the thing about the law of sines. I'm going to try to prove the HL theorem and maybe I could get some insight on it.
Directrix
  • Directrix
Sounds like a good plan.

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