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Similar triangles, looking for demonstration

  1. Aug 28, 2005 #1
    I need an algebric proof of the theorem of similar triangles (C/c=A/a=B/b).
  2. jcsd
  3. Aug 28, 2005 #2


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    Proof of what exactly?
  4. Aug 28, 2005 #3
    For two similar triangles, that A/a=C/c=B/b.
  5. Aug 28, 2005 #4


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    I take it you mean the sides with A,B,C and a,b,c?

    Your proof would depend on who you define similar triangles, I assume that's saying that the corresponding angles of both triangles are the same.
  6. Aug 28, 2005 #5
    wouldnt the proof be something like this

    Sin A / a = Sin B /b = Sin C / c
  7. Aug 28, 2005 #6
    Well trigonometric functions don't proove the theorem, they are a product of it so it is not really a proof... Anyway I'm putting my mind into it and I hope to find why it is so.
  8. Aug 28, 2005 #7


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    Your problem is actually pretty straightforward but the question I have for you is "What have you tried so far?" This sounds like a homework problem.
  9. Aug 29, 2005 #8
    I was finally able to proove it. Here is my solution, but perhaps there is simplest one and I would like someone to show it...

    1.Two similar right triangles of sides a,b,c and x,y,z.

    The following is true:

    c^2 + z^2 + 2cz=a^2 + x^2 + 2xa + b^2 + y^2 + 2by
    Since c^2=a^2 + b^2, and z^2=x^2 + y^2,
    2cz=2xa + 2by
    cz=xa + by
    For simplifying purpose, square the equation;

    (cz)^2=(xa + by)^2
    (cz)^2=(xa + by)^2

    (x^2 + y^2)(a^2 + b^2)=(xa + by)^2

    (ax)^2 + (xb)^2 + (ay)^2 + (by)^2=(ax)^2 + (by)^2 + 2xaby


    (xb)^2 + (ay)^2 - 2xaby = 0

    Factorising, we find that


    To proove that z/c=x/a=y/b,


    (z/c)^2=(x^2 + y^2)/(a^2 + b^2);
    (z/c)^2=(x^2 + (bx/a)^2)/(a^2 + (ay/x)^2)
    (z/c)^2=(x^2(1+(b/a)^2)/(a^2((y/x)^2 + 1)

    Since we know that b/a=y/x, we obtain the result

    Since any triangle can be decomposed into two right triangles, it is easy to prove that the rule is general.
    Last edited: Aug 29, 2005
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