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Math Help - Angles proving

  1. #1
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    Angles proving

    Point M is the middle of a BC side of an acute triangle ABC. Point K lies on the BC side and fills the condition \angle BAM=\angle KAC. On the AK segment such point E was chosen that \angle BEK=\angle BAC. Prove that \angle KEC = \angle BAC.

    From what see in my provisory image, the M point is the same point as the K one since KAC divides the BAC angle in the half and thus for BAM to be the other half, M must be K. Then, we choose an E on AK and since we know that AK (or AM, it's the same) divides BAC in the half, automatically BEK = KEC and since we've selected such E that BEK=BAC, KEC=BAC.

    Is my thinking right? Cause frankly - I don't think so..
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  2. #2
    MHF Contributor Unknown008's Avatar
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    I don't think it's wrong... Seems good to me.
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  3. #3
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    Hmm.. After second thought I see a problem in the first sentence of my solution. For M to be the same as K, AK would have to be the bisector of BAC and the median of BC at the same time which it rather isn't, is it?
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Hm.. yes, it's only the case if triangle ABC is isosceles with sides AB and AC equal. I had an equilateral triangle as sketch and hence, it automatically worked...

    I tried another drawing and I can't find a simple way to prove this.
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  5. #5
    MHF Contributor Unknown008's Avatar
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    Hey, Opalg solve it here. It's the exact same problem.

    I don't think I would have been able to do this. Really interesting problem
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