How to prove this geometry problem mathematically? Triangle ∆ABC is arbitrary. Side AC is a diameter to a semicircle, and M marks the point in the middle of this semicircle arc. Side BC is a diameter to a semicircle, and N marks the point in the middle of this semicircle arc. O is a point on side AB, and AO = OB.
Conjecture: OM = ON  and $\measuredangle$MON = 90°
 A: With the vertices represented by complex numbers $a,b,c$, we find $m=a+\alpha(c-a)$, $n=c+\alpha(b-c)$, $o=\frac{a+b}2$ with $\alpha=\frac{1+i}2$.
Now oberve that 
$$m-o = -\frac i2a-\frac12 b+\frac{1+i}2c$$
and 
$$n-o = -\frac 12a+\frac i2 b+\frac{1-i}2c$$
and hence indeed $ m-o = i(n-o)$.
A: Let's suppose two exterior semicircles in our approach. See the figure bellow.

Note that CJOK is a paralelogram, that $\angle CJO$ is external of $\triangle JAO$ and that $\angle CKO$ is external of $\triangle KOB$.
Note also that $\triangle MJO$ and $\triangle OKN$ are congruent, therefore 
$$MO = ON.$$
Using $\triangle MJO$ we get:
$$\alpha + \beta + 90^{\circ} + \theta + \gamma = 180^{\circ} \Rightarrow$$
$$\alpha + \beta + \theta + \gamma = 90^{\circ}$$
But as $\angle AOB$ is a straight angle and $\alpha + \beta + \theta + \gamma = 90^{\circ}$ we can conclude that $\angle MON$ is a right angle.
You can use a similar approach when semicircles are not exterior.
A: By "mathematically" do you mean by coordinates as opposed to geometry?  By coordinates, you can let $A=(-1,0), B=(x,y), C=(1,0)$ by rotation, translation, and scaling.  Then $M=(0,-1), N= (\frac{x+y+1}2,\frac{x+y-1}2)O=(\frac x2,\frac y2)$.  Now you can check.
