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 . On the AK segment such point E was chosen that . Prove that .
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.. :)
Oct 16th 2010, 09:47 AM
I don't think it's wrong... Seems good to me. (Smile)
Oct 19th 2010, 06:29 AM
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?
Oct 20th 2010, 06:27 AM
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. (Doh)
Oct 27th 2010, 07:53 AM
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 (Smile)