Using the Law of CosinesDate: Fri, 18 Aug 1995 17:24:28 +1000 From: Tony Snell Subject: Mathmatical proof, cosine rule G'day, I have been given this problem solving project. Prove that, in any triangle , the sum of the squares of two sides is equal to twice the square on half the third side plus twice the square on the line joining the midpoint of the third side to its opposite vertex (Hint: You may find the cosine rule useful). So far I have this equation : a^2 + c^2 = 2(x/2) + (b^2/2) . My teacher said this was correct. x = the line in the middle of the large triangle. a = the line opposite the angle A b = the line opposite the angle B c = the line opposite the angle C The teacher kind of hinted in class that I should use a simultaneous equation to cancel out something. I know have to prove the left hand side equals the right hand side of the equation and use a variant of the cosine rule to prove the equation, but how? I hope you have and idea. Thanks, Marc Snell. From: Doctor Ken Subject: Re: Mathmatical proof, cosine rule Hello Tony! I've got a question about your first equation. When you said "twice the square on the line joining the midpoint of the third side to its opposite vertex," it seems like that would translate to 2(x^2) instead of 2(x/2) like you have written. Right? Anyway, let's use the Law of Cosines: a^2 = b^2 + c^2 - 2bcCos(A) Now try to use the Law of Cosines again, on one of the smaller triangles, the one that contains angle A. Hint: you're trying to get that Cos(A) term to go away, so see if you can make your next Law of Cosines equation contain that term, then use simultaneous equations (with the original equation) to get rid of that term altogether. Good luck! -Doctor Ken The Geometry Forum |
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