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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.

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    $\begingroup$ Hint: show that $\angle AHC=180^\circ-\angle ABC$. $\endgroup$ – user10354138 Feb 6 at 18:18
  • $\begingroup$ 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? $\endgroup$ – Enc_23 Feb 6 at 18:28
  • $\begingroup$ Essentially the same thing. $\endgroup$ – player3236 Feb 6 at 18:38
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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.

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  • $\begingroup$ 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? $\endgroup$ – Enc_23 Feb 6 at 18:55
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    $\begingroup$ 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. $\endgroup$ – dodoturkoz Feb 6 at 19:10
  • $\begingroup$ Thank you, I finally got it. It was easier than I thought. I was actually missing the obvious. $\endgroup$ – Enc_23 Feb 6 at 19:28
  • $\begingroup$ Answering questions where the asker has shown no effort at all often attracts downvotes, because this incentivises these askers to write more questions without effort since they can get an answer. Your answer to this question was deleted because of the downvotes, which put it into the delete queue. $\endgroup$ – Toby Mak Feb 11 at 8:08
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    $\begingroup$ @TobyMak I feel like the heavy moderation of this website increased my reluctance to write answers in a topic that I’m very passionate about, geometry. At least, thank you for sparing your time to explain it. $\endgroup$ – dodoturkoz Feb 11 at 8:39
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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.

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  • $\begingroup$ This is a more beautiful proof but harder to notice if this property is not seen before. $\endgroup$ – dodoturkoz Feb 6 at 19:15
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    $\begingroup$ @dodoturkoz That's right! That's also the reason why I upvoted your answer (which might definitively be more useful for the OP), but could not resist to post this approach — which, in turn, might be helpful for other users :) $\endgroup$ – Dr. Mathva Feb 6 at 19:27

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