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Let $\triangle ABC$ be acute, $H$ is its orthocenter, and $(C)$ is its circumcircle. $E,F,D$ are the orthogonal projections of $H$ on $AC,AB,BC$. Let $H'$ be the reflection of $H$ about $BC$.

Prove that $H'\in (C)$.

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I don't know if that's right but I think that if we can show that $ABH'C$ is convex then we're done, right?

Note that $\triangle EBC$ and $\triangle ACD$ are both right triangles and they share $\angle C$, which means $\angle CAD=\angle EBC$. Also in $\triangle HDB$ and $\triangle BDH'$, we have $HD=DH'$ and they're both right triangles, so $\angle HBD=\angle DBH'$. Hence $CAH' = CBH'$, thus $ABH'C$ is cyclic. Is that correct?

Jean Marie
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PNT
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  • https://math.stackexchange.com/questions/589877/why-would-the-reflections-of-the-orthocentre-lie-on-the-circumcircle#:~:text=Since%20points%20B%20and%20Q,the%20circumcircle%20of%20that%20triangle. – Math Lover Feb 05 '22 at 11:40
  • Also, your work is correct. By showing that $\angle CAH' = \angle CBH'$, you have already shown that $ABH'C$ is cyclic. – Math Lover Feb 05 '22 at 11:44
  • Your title is erroneous: reflection is wrt sides $AB,BC,CA$... – Jean Marie Feb 06 '22 at 21:13

1 Answers1

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Yes. Because Angles in the same segment are equal and the converse is also true. So, if $\angle CAH' = \angle CBH'$, the $B$ must lie on the circle passing through $C, A $ and $H'$. In other words, $H'$ lies on the circumcircle of $\Delta ABC$

Here is a more direct proof of $ABCH'$ being a cyclic quadrilateral.

You showed that $\angle CAH' = \angle CBH'$. Similarly you can show $\angle BAH' = \angle BCH'$.

We know that
$$\angle BCH' + \angle CH'B + \angle H'BC = 180$$ $$\implies \angle BAH' + \angle CH'B + \angle H'AC = 180$$ $$\implies \angle H'AB + \angle H'AC + \angle CH'B = 180$$ $$\implies \angle BAC + \angle BH'C = 180$$

Which means, $ABCH'$ is a cyclic quadrilateral.

Chaitanya Chavali
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  • But if $H'\in (C)$ and $H'BHC$ is convex then that means $H\in (C)$?? – PNT Feb 05 '22 at 11:27
  • @Yassir Sorry. There was a typo. I edited the answer. $H'BHC$ is convex because it's an acute triangle, but that doesn't still mean they are cyclic. cyclic $\implies$ convex, but converse is not true. – Chaitanya Chavali Feb 05 '22 at 12:08