# Why radiuses of the circumcircles are equal? [closed]

Why radiuses of the circumcircles of triangles ABC, AHC, BHC and ABH are equal? I know it can be proven using the law of sines, but I can't figure how.

• Hint: show that $\angle AHC=180^\circ-\angle ABC$. – user10354138 Feb 6 at 18:18
• Thanks. Then $sinAHC=sinABC$ and $R=\frac{AC}{2\sin \left(ABC\right)}$ for $\triangle ABC$ and $\triangle AHC$. But what to do with the other two triangles? – Enc_23 Feb 6 at 18:28
• Essentially the same thing. – player3236 Feb 6 at 18:38

As one of my favorite lecturers would say, "Triangle is a democratic figure, if you apply a theorem for one vertex, you should apply it for the others too." $$180^\circ-m(\angle ABC)=m(\angle AHC)$$ $$180^\circ-m(\angle BCA)=m(\angle BHA)$$ $$180^\circ-m(\angle BAC)=m(\angle BHC)$$ Therefore, all sine values are equal, implying the equality of circumradii.
• I'm probably missing something really obvious. $\angle AHC = \angle KHM, \angle ABC = 360-90\cdot 2 - \angle KHM = 180-\angle ABC$ so I proved it with KBMH. I can't do the same thing with $\angle BHA$ and $\angle BHC$. Is there a different way? – Enc_23 Feb 6 at 18:55
• You can do the exact same thing. Extend $BH$ and note that it will be perpendicular to $AC$. The reason is that the three altitudes of a triangle meet at one point (at the orthocenter), so $BH$ is also an altitude of the triangle. – dodoturkoz Feb 6 at 19:10
Hint. Prove (say: with angle chasing) that the reflections of the orthocenter across each side $$AB, BC, CA$$ respectively lie on the circumcircle of $$\triangle ABC$$. Since the resulting triangles $$H_cAB, \ldots$$ are congruent to the original ones, you’re done.