Note that by symmetry, it suffices to show that $AA_2, BB_2$ and $EE_2$ are concurrent. To show this, it suffices by the Radical Axis Theorem to prove that $BB_2E_2E$ is cyclic. Let $\Omega_e, \Omega_b$ denote $(\triangle AE_2D), (\triangle AB_2C),$ respectively. Let $A_3 = \Omega_e \cap \Omega_b$, $X = BD \cap \Omega_e$, and $Y = CE \cap \Omega_b.$ Notice that $\angle DA_3A = 180 - \angle AE_2D = \angle E_2AC_1 + \angle E_2DC_1 = \angle E_2EC_1 + \angle E_2EB_1 = \angle BEC.$ Analogously, $\angle CA_3A = \angle DBE.$ Therefore, $\angle DA_3C = \angle DA_3A + \angle CA_3A = \angle BEC + \angle DBE = 180 - \angle DA_1C,$ and so $A_1CA_3D$ is cyclic. Observe that $\angle BXE_2 = \angle DAE_2 = \angle C_1AE_2 = \angle C_1EE_2 = \angle BEE_2,$ and so $XBE_2E$ is cyclic. Analogously, $YEB_2B$ is cyclic. Notice that: $$\angle XAD = \angle XE_2D = 180 - \angle E_2XD - \angle E_2DX = 180 - \angle E_2AD - \angle E_2DB = 180 - \angle E_2EC_1 - \angle E_2DA - \angle ADB = 180 - \angle E_2EC_1 - \angle E_2EB_1 - \angle ADB = 180 - \angle BEC - \angle ADB.$$ Hence, it follows analogously that $\angle YAC = 180 - \angle DBE - \angle ACE.$ Hence, we get that $\angle XAY = \angle XAD + \angle YAC - \angle DAC = 180 - \angle BEC - \angle ADB + 180 - \angle DBE - \angle ACE - \angle DAC = 180$, i.e., $X, A, Y$ are collinear. Therefore, we know that $\angle YXB = \angle AXD = 180 - \angle AE_2D = \angle E_2AD + \angle E_2DA = \angle BEC = 180 - \angle BEY$, and so it hence follows that $YXBE$ is cyclic. Since $XBE_2E, YEB_2B, YXBE$ are all cyclic, it easily follows that $XBB_2E_2EY$ is cyclic, and so hence $BB_2E_2E$ is cyclic as desired. $\square$

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I see the similar problem. Please see the figure, $A$, $B$, $C$, $D$, $E$ are any points on plane. Prove that five red circles have same radical center. I think that this problem is different with Telv's because, in my configuration, the red circles don't have any common point before?

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Let $AA_3$ meets $\odot(A_3C_1D_1)$ again at $A_4$. By Reim's Theorem, $D_1A_4 \parallel EA$ and $C_1A_4 \parallel AB$. Hence $A_4$ can be defined as $A_4 = C_1\infty_{AB} \cap D_1\infty_{AE}$. Define $B_4, C_4, D_4, E_4$ similarly, it suffices to show that $AA_4, BB_4, CC_4$ are concurrent. I believed that it simplify the problem, however, I haven't come up with an geometric proof. But found out that it's not too complicated while using bary bashing. So let do it! Let $A = (0:1:0), B=(1:0:0), C = (0:0:1)$ and $D = (u:v:w), E=(p:q:r)$. Then \[E_1 = (0:v:w) \qquad D_1 = (0:q:r)\]Since the infinite point of $BC, BA$ is $(1:-1:0)$ and $(1:0:-1)$, respectively. We have \[ \begin{aligned} E_1\infty_{AB}: & \quad 0 = wx+wy-vz \\ D_1\infty_{BC}: & \quad 0 = qx-ry+qz \end{aligned} \]So we have \[ B_4 = ( - :vq+wq:wq+wr) \]Now we compute $A_4$. Note that the formula of $AD$ and $BE$ is $(u:-:w)$ and $(-:q:r)$, respectively. Hence \[C_1 = (ur:qw:rw)\]And also $\infty_{AE} = (p:-p-r:r)$, so we have \[ \begin{aligned} D_1\infty_{AE}: & \quad 0 = r(p+q+r)x +rpy -pqz \\ C_1\infty_{AB}: & \quad 0 = rwx +rwy -(ur+qw)z \\ \end{aligned} \]Hence \[ \begin{aligned} A_4 &= (pq\cdot w - (ur+qw) \cdot p: - :r(p+q+r)\cdot w - rw \cdot p) \\ &=(-urp: - :rw(q+r)) \\ &=(-up:-:w(q+r)) \end{aligned} \]Similarly, \[ C_4 = (-up:q(v+w):-) \]So eventually, that $AA_4, BB_4, CC_4$ are concurrent becomes an obvious results.

It’s just Geometry in figures 2.38 (second edition) or 2.37(first edition).

Nice problem! Sketch: Define $K,L = (AE_2D) \cap \overline{BD},(AB_2C) \cap \overline{CE}$, by little angle chase $BKLE$ is cyclic, next $BB_2EL$ is cyclic and similarly $EE_2BK$ is cyclic. So $BB_2EE_2$ is cyclic and so by radax we're done.