So close, but no cigar

Hi Noobmaster69,

Let O and H be the circumcenter and orthocenter of acute triangle ABC, respectively.

If the sum for the degrees of angle AHC  (orange) and angle AOC (pink) is 240 degrees

what is the measure of angle ABC  in degrees?   (black)

Here is the pic

The  pink plus orange angles add up to 240 degrees and I have to find the size of the black angle.  

My pic in not to scale, I mean my pink and orange ones do not add up to 240. 

Other than that I think my pic is a proper indication of the problem at hand.

Let the black angle be called alpha       \(\angle CBA=\alpha\)

Consider minor arc CA

< COA is the angle at the centre standing on arc CA    ....  that is the pink angle

< CBA  is the angle at the circumferecne standing on arc CA    ....  that is the black  angle 

so since the angle at the centre = 2 times the angle at the circumference when both are staning on the same arc,  

BLACK    \(\angle CBA=\alpha\)

Pink         \(\angle COA=2\alpha\)

NOW  look at the new image below.

Consider the green triangle \(\triangle IBA\) and the yellow triangle \(\triangle KHA\)

angle AIB = angle AKH = 90 degrees

angle at A is a common angle

therefore these two triangles are similar which means that   \(\angle IBA=\angle KHA = \alpha\)

Now I have 

\(\angle CHA+\angle KHA = 180\\ \angle CHA+\alpha = 180\\ orange + \alpha =180 \;degrees\\ orange \;angle=(180-\alpha) \;degrees\\\)

I also know that

 \(orange+pink=240\\ (180-\alpha)+(2\alpha)=240\\ 180+\alpha=240\\ \alpha=60\;degrees\)

Therefore, angle ABC=60 degrees

I know my logic is correct but i hope i have not made a mistake during the presentation.

Plus

There may well have been a much simpler way to do it.

YIKES!    I had no inclination....or know-how .....to answer THIS one !

Nice , Melody !