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Let $H$ be the orthocentre of an acute $\Delta ABC$, as in figure. Let $X$ be the reflection of $H$ over $\overline{BC}$ and $Y$ be the reflection of $H$ over the midpoint of $\overline{BC}$. Show that $X$ and $Y$ lie on the circumcircle of $\Delta ABC$.

To prove that $X$ and $Y$ lie on the circumcircle of $\Delta ABC$, I tried showing that $AXYC$ and $ABXY$ are both cyclic which would make $A, B, C, X, Y$ concyclic and since the circumcircle is unique for a given triangle, $X$ and $Y$ must lie on it (follows from the points being concyclic).

I doubt this is sufficient as a proof for this problem. Can someone guide me? I would like to continue with this proof unless it's not convenient to prove by these arguments.

amWhy
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Harshul
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  • Welcome to Math.SE! ... This ancient answer discusses the case of $X$. From here, handling $Y$ might be pretty straightforward. – Blue Mar 10 '24 at 12:46
  • Please, choose the tags carefully. This shouldn't be tagged "solution-verification" – jjagmath Mar 10 '24 at 12:52
  • @Blue oh! This was much easier. I think I can handle $Y$ from there. Thank you very much. Is it possible for you to see if my argument is sufficient for the problem? Thanks again. – Harshul Mar 10 '24 at 12:55
  • @jjagmath I'm sorry. I will remove that tag. – Harshul Mar 10 '24 at 12:56
  • For the $X$ part, do some simple angle chasing. For the $Y$ part, simple angle chasing + finding a parallelogram should help. As an additional practice, try proving $AY$ is the circumdiameter. – D S Mar 10 '24 at 13:01
  • @DS Yes, the parallelogram $BHCY$. I used that to prove $ABXY$ is cyclic. Is it enough to say that $\overline{AY}$ is the circumdiameter because $\angle AXY$ is a right angle (as $\Delta HDE \sim \Delta HXY$ where $D$ is the foot of perpendicular of $A$ on $\overline{BC}$ and $E$ is the midpoint of $\overline{BC}$)? – Harshul Mar 10 '24 at 13:07
  • Yes, so what is your question exactly? – D S Mar 10 '24 at 13:09
  • I tried showing that $AXYC$ and $ABXY$ are both cyclic which would make $A, B, C, X, Y$ concyclic and since the circumcircle is unique for a given triangle, $X$ and $Y$ must lie on it (follows from the points being concyclic). However, I doubt this is sufficient as a proof for this problem. Is it? – Harshul Mar 10 '24 at 13:12
  • Yes it is sufficient, but difficult. Instead, you can prove ABXC and ABYC, why don't you go along with that? – D S Mar 10 '24 at 14:07
  • I could definitely do that. But I have already proved $AXYC$ and $ABXY$ are cyclic by a little bit of angle chasing. Thank you very much for your help. – Harshul Mar 10 '24 at 15:22

1 Answers1

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I will take it that it is already established that points $A$, $C$, $X$ and $B$ are concyclic. Now, we shall construct the perpendicular bisector of $BC$, which we will call $D$. Reflecting points $H$ and $X$ around line $D$ give $H'$ and $Y$ respectively. (I will leave you to prove that reflecting $X$ around line $D$ gives $Y$. You will need to use $H'$. )

As $BC$ is a chord of the circumcircle of $\triangle ABC$, its perpendicular bisector $F$ is a diameter of the circle. Thus, $A$, $B$, $C$, $X$ and $Y$ are concyclic. (The reflection of a point lying on the circumference of a circle upon its diameter lies on that circle itself.)

Matthan
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  • Just let me know if you cannot prove that reflecting X around line D gives Y. I can help you. – Matthan Mar 12 '24 at 04:50
  • Is it right to say that reflecting $X$ around line $D$ gives $Y$ because $\Delta HEM \sim \Delta HXY$ where $M$ is the midpoint of $\overline{BC}$ and $E$ is the foot of perpendicular of $A$ on $\overline{BC}$. Moreover, $\Delta HXY$ is scaled by a factor of $2$. Since $M$ is already the midpoint, it should imply that $Y$ is the reflection about line $D$. – Harshul Mar 13 '24 at 00:49
  • Yes, about there. I was thinking of further constructing a rectangle $HH'YX$, but you have shown that it is unnecessary. – Matthan Mar 13 '24 at 08:30
  • I can see how this would work out. Thank you very much for your help. I will accept this answer. – Harshul Mar 13 '24 at 11:54