# Angles proving

• Oct 16th 2010, 09:13 AM
Glyper
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 \$\displaystyle \angle BAM=\angle KAC\$. On the AK segment such point E was chosen that \$\displaystyle \angle BEK=\angle BAC\$. Prove that \$\displaystyle \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.. :)
• Oct 16th 2010, 09:47 AM
Unknown008
I don't think it's wrong... Seems good to me. (Smile)
• Oct 19th 2010, 06:29 AM
Glyper
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
Unknown008
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
Unknown008
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)