By an easy angle chasing, we see that both conditions are equivalent to $\angle BAD =\angle CAE$.

If we invert at $A$ and consider a homothety at $A$ with scale $1/2$ (for the centers $O_1,O_2,\dots,O_6$ only), then our problem would become equivalent to the following (we will also be using the fact that points $O_1,O_3,O_4,O_5$ don't lie on a line, which is true for obvious reasons): Inverted Problem wrote: Let distinct points $A,B,D,E,C$ lie on a circle in that order. Let $O_1,O_2,O_3,O_4,O_5,O_6$ be the projections of $A$ onto segments $BD,DE,EC,BE,DC,BC$, respectively. Prove that points $O_1,O_3,O_4,O_5$ are concyclic if and only if points $A,O_2,O_6$ are collinear. [asy][asy] size(250); pair A=dir(100),B=dir(-160),C=dir(-20),D=dir(-130),E=dir(-50); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); draw(unitcircle); pair O1=foot(A,B,D),O2=foot(A,D,E),O3=foot(A,E,C),O4=foot(A,B,E),O5=foot(A,D,C),O6=foot(A,B,C); dot("$O_1$",O1,dir(O1)); dot("$O_2$",O2,dir(O2)); dot("$O_3$",O3,dir(O3)); dot("$O_4$",O4,dir(O4)); dot("$O_5$",O5,dir(O5)); dot("$O_6$",O6,dir(O6)); draw(O1--D--E--O3,magenta); draw(B--C--D,magenta); draw(B--E,magenta); draw(O1--O2--O3,royalblue); draw(O1--A--O2,red); draw(O4--A--O5,red); draw(A--O3,red); draw(B--A--C,magenta); [/asy][/asy] By Simson line, we have that lines $O_1O_4,O_3,O_5$ concur at $O_2$. Now, let $\omega_1,\omega_2$ be the circle with diameter $\overline{AB},\overline{AC}$, respectively. Then $O_1,O_4 \in \omega_1$ and $O_3,O_5 \in \omega_2$. Now, with directed lengths we have \begin{align*} & \text{points } O_1,O_3,O_4,O_5 \text{ are concyclic} ~ \iff ~ O_2O_4 \cdot O_2O_1 = O_2O_5 \cdot O_2O_3 \\ & \iff ~ \text{Pow}_{\omega_1}(O_2) = \text{Pow}_{\omega_2}(O_2) ~ \iff ~ O_2 \text{ lies on the radical axes of } \omega_1,\omega_2 \end{align*}But the radical axes of $\omega_1,\omega_2$ is just the line $AO_6$. Hence, $$\text{points } O_1,O_3,O_4,O_5 \text{ are concyclic} ~ \iff ~ O_2 \in \overline{AO_6} ~ \iff ~ \text{points } A,O_2,O_6 \text{ are collinear} $$This completes the proof of the problem. $\blacksquare$