Reflections meet on circumcircle of ABC
- Enthurelxyz
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Let $l$ is the Euler line of $\triangle ABC$. $l_a,l_b,l_c$ are the reflections of $l$ from $BC,AC,AB$ respectively. Prove that $l_a,l_b,l_c$ meet on the circumcircle of $ABC$.
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- Anindya Biswas
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Re: Reflections meet on circumcircle of ABC
Let $O$ and $H$ be the circumcenter and orthocenter of $\triangle ABC$ respectively.
Let $X, Y, Z$ be the intersection point of pairs of lines $(l, l_a), (l, l_b), (l, l_c)$ respectively.
Since $l, l_a$ are reflections with respect to line $BC$, so the line $BC$ must go through $X$. Similarly, the lines $CA, AB$ goes through the points $Y,Z$ respectively.
It is a well known fact that the reflection of the orthocenter with respect to a side of the triangle lies on the circumcircle of that triangle.
Let $D,E,F$ be the reflections of $H$ with respect to $BC, CA, AB$ repectively. So, $D,E,F$ must lie on the circumcircle of $\triangle ABC$. Also, $D,E,F$ lies on $l_a, l_b, l_c$ respectively since $H$ lies on $l$.
Let $P$ be the intersection point of pairs of lines $(l_b, l_c)$. Let $\measuredangle (l_b,l_c)$ be the directed angle mod $\pi$ that rotates the line $l_b$ to $l_c$.
Now, $\measuredangle (l_b,l_c)=\measuredangle EPF=\measuredangle YPZ=-\measuredangle PZY-\measuredangle ZYP$
$=\measuredangle YZF+\measuredangle EYZ=2(\measuredangle YZA+\measuredangle AYZ)=2\measuredangle YAZ=2\measuredangle CAB=\measuredangle EAF$.
So, $EPFA$ is a cyclic quadrilateral. So, $P$ lies on the circumcircle.
Let $Q$ be the intersection point of pairs of lines $(l_c, l_a)$. By the similar argument above, we can show that $Q$ also lies on the circumcircle.
So, $l_c$ intersects the circumcircle at $F, P, Q$. Since a line can only intersect a circle on at most $2$ distinct points, either $Q=P$ or $Q=F$.
If $Q=F$, then $l_a$ would be the line joining $F,D$. In this case, $l_a, l_b$ both has $2$ intersection points with the circle and non of them are intersection of $l_a$ and $l_b$. But this contradicts the fact that $(l_a, l_b)$ must have a intersection point on the circle.
So, it must be the case that $Q=P$. That means, $l_a, l_b, l_c$ goes through $P$ which lies on the circumcircle of $\triangle ABC$.
Q.E.D.
Let $X, Y, Z$ be the intersection point of pairs of lines $(l, l_a), (l, l_b), (l, l_c)$ respectively.
Since $l, l_a$ are reflections with respect to line $BC$, so the line $BC$ must go through $X$. Similarly, the lines $CA, AB$ goes through the points $Y,Z$ respectively.
It is a well known fact that the reflection of the orthocenter with respect to a side of the triangle lies on the circumcircle of that triangle.
Let $D,E,F$ be the reflections of $H$ with respect to $BC, CA, AB$ repectively. So, $D,E,F$ must lie on the circumcircle of $\triangle ABC$. Also, $D,E,F$ lies on $l_a, l_b, l_c$ respectively since $H$ lies on $l$.
Let $P$ be the intersection point of pairs of lines $(l_b, l_c)$. Let $\measuredangle (l_b,l_c)$ be the directed angle mod $\pi$ that rotates the line $l_b$ to $l_c$.
Now, $\measuredangle (l_b,l_c)=\measuredangle EPF=\measuredangle YPZ=-\measuredangle PZY-\measuredangle ZYP$
$=\measuredangle YZF+\measuredangle EYZ=2(\measuredangle YZA+\measuredangle AYZ)=2\measuredangle YAZ=2\measuredangle CAB=\measuredangle EAF$.
So, $EPFA$ is a cyclic quadrilateral. So, $P$ lies on the circumcircle.
Let $Q$ be the intersection point of pairs of lines $(l_c, l_a)$. By the similar argument above, we can show that $Q$ also lies on the circumcircle.
So, $l_c$ intersects the circumcircle at $F, P, Q$. Since a line can only intersect a circle on at most $2$ distinct points, either $Q=P$ or $Q=F$.
If $Q=F$, then $l_a$ would be the line joining $F,D$. In this case, $l_a, l_b$ both has $2$ intersection points with the circle and non of them are intersection of $l_a$ and $l_b$. But this contradicts the fact that $(l_a, l_b)$ must have a intersection point on the circle.
So, it must be the case that $Q=P$. That means, $l_a, l_b, l_c$ goes through $P$ which lies on the circumcircle of $\triangle ABC$.
Q.E.D.
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- Anindya Biswas
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Re: Reflections meet on circumcircle of ABC
Another remark : I only used the fact that $l$ goes through $H$. Any line going through $H$ has this nice property, not just the very very special Euler Line!
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
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- Mehrab4226
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Re: Reflections meet on circumcircle of ABC
I didn't read your proof yet, Aninda Biswas. So it might be the same. If so then sorry
In the figure, the pink line is $l_C$ and the orange line is $l_B$
![Sad :(](./images/smilies/icon_e_sad.gif)
In the figure, the pink line is $l_C$ and the orange line is $l_B$
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