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?enter image description here


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 $ABC$.

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  • $\begingroup$ If three circles are equal and passing through H ..how that means that distance from H to A'B' , B'C' ,C'A' are the same.......I think I am missing something ...can you pls provide the missing details.. $\endgroup$ – IMO 2021 GOLD Mar 25 at 4:36
  • $\begingroup$ @User12002 I've added an explanation. $\endgroup$ – Carl Schildkraut Mar 25 at 6:51

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.

enter image description here

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  • $\begingroup$ I have 2 doubts in your proof ..1) if radius is equal of both circles then how they map to lines which are equal distance from C ...2) how HBC=B'HC did uu apply angle segment theorem ? Can yu pls give more details to your answer ... $\endgroup$ – IMO 2021 GOLD Mar 24 at 3:00

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