Prove that H is orthocentre of ABC using inversion 
Three equal circles pass through a given point $H$ and meet one another two by two at $A,B,C$ prove that $H$ is orthocentre of triangle $ABC$.

My try - 
I proved it using elementary geometry methods quite easily but i also want to prove it using inversion...
so first i invert about H to make all three circles lines but not able to go further to prove that it is orthocentre of ABC...
Any hints?
 A: Invert at $H$. The fact that the three circles through $H$ are equal means that, if $A',B',C'$ are the images of $A,B,C$, then the distances from $H$ to $A'B'$, $B'C'$, and $C'A'$ are the same, because the point diametrically opposite $H$ on $(HAB)$, for example, inverts to the foot from $H$ to $A'B'$. Therefore $H$ is the incenter of $A'B'C'$. Now, let $X,Y,Z$ be the $A'$-, $B'$-, and $C'$-excenters of $A'B'C'$. We see that $H$ is the orthocenter of $XYZ$ and that a negative inversion about $H$ sends $A'\to X$, etc., so $ABC$ is directly similar at $H$ to $XYZ$, and so $H$ is the orthocenter of $ABC$. 
A: Observe an inversion with center at $C$ and radius $r= CH$. 
Let $A\mapsto A'$ and $B\mapsto B'$. 
Since circles (Y) and (Z) have equal radius and go through $C$ they map to a lines $CA'$ amd $CB'$ which are at equal distance from $C$. So $HC$ is angle bisector for $\angle A'HB'$. Now we have $$\angle HBC = \angle B'HC = \angle A'HC = \angle HAC =:\alpha $$
By simmetry we have $$\angle HBA = \angle HCA =:\beta $$ and $$\angle HAB = \angle HCB =:\gamma $$
Now it is clealry $$\alpha +\beta +\gamma = 90^{\circ}$$ so we are done.

