Math contest - Geometry

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Hey all,

 

so, I was at this math contest last Tuesday. The contest consisted of 4 question, of which one was geometry. I had absolutely no clue on how to solve it, so I was wondering if anybody here could help me.

 

We have a triangle $ABC$, which is not isosceles. Let $x$ be a line that bisects the angle $ACB$, $y$ be the perpendicular bisector of $AB$, $h_a$ be a line perpendicular to $BC$ and passing through $A$ and finally $h_b$ be a line perpendicular to $AC$ and passing through $B$. Let $K \in x \cap y$, $P \in KC \cap h_a$ and $Q \in KC \cap h_b$. Let $A_{AKP} = A_{BKQ}$, where $A_{XYZ}$ denotes the area of the triangle $XYZ$. Determine the size of the angle $ACB$.

 

Since my knowledge of English mathematical terms is unsatisfactory at best, I attached a picture in case anyone was confused. Anyway, I already know the correct answer is 60°, but I haven't got the faintest idea why. I don't need the result for anything anymore, but I sure am curious, so thanks in advance for any light shed on this matter.

 

Cheers,

 

Gabe