Let $P$ be the orthocenter of $\triangle ABC$, and let $A'$ be the reflection of $A$ about $\overline{BC}$. Show that $A'$, $B$, $C$, and $P$ are concyclic (i.e. lie on the same circle).
Maybe this is easy, and it definitely seems to be true based on empirical evidence, but can anyone supply a proof of this? I am not an expert in Euclidean geometry, so my apologies if this is known/trivial!
The issue seems to be the fact that a proof may depend on the picture. Well, let us consider first the "standard" picture. The orthocenter is $H$ below, so that the hand will not type against my will. A prime denotes a reflection w.r.t. the line $BC$. The points $A',B,C,H$ are concyclic if and only if their reflections $A''=A,B,C,H'$ are so. We have thus to show that $H'$ is on the circumcenter of $\Delta ABC$. In the "standard" picture:
$$ \begin{aligned} \widehat{BAH'} &= \widehat{HCB} &&\text{ their complement is $B$,}\\ &=\widehat{BCH'} \ ,&&\text{ so $A,B,C,H'$ concyclic.} \end{aligned} $$ What about other positions of the points? In case $A$ is obtuse, then $A,H$ change their places in the figure, so we have the same problem with points to be concyclic in an other order.
In case $B$ is obtuse we have the following picture:
and the same argument applies. Instead of the complement of $B$ we use the complement of $180^\circ-B$. (Or we use the correct angles in green, which are not the ones marked in green, but their "opposite versions", opposite in the sense of $x\to360^\circ -x$.)
In case $C$ is obtuse, switch $B$ and $C$ to go back to the previous case.