Let $H$ be the orthocenter of the triangle $\Delta ABC$ and point $A_1$ be a reflection of $H$. Prove $A_1$ belongs to the circumscribed circle of $ABC$.
My attempt:
Let $AH\perp BC$ and $D\equiv AH\cap BC$ and $E\equiv AB\cap EH$. $$\implies |HD|=|DA_1|\implies \Delta HA_1C\;\text{is isoceles}$$ $$HD\perp BC\;\land\;BE\perp EH\implies\measuredangle EBC=\measuredangle CHD=\measuredangle HA_1C$$ $\implies$ $\measuredangle EBC=\measuredangle AA_1C$ are inscribed angles of the circumscribed circle. $\implies A_1$ is on the circumscribed circle.
Picture: 
Is this legitimate and is there a more efficient method? How can I improve my proof?
Thank you in advance!
Your method is correct. If you are familiar with the Euler line and dilation, you can see a dilation between those points (D and A1) and between N (the nine point center) and O (the circumcenter) from H(the orthocenter).