Let $I$ the incenter of $\triangle ABC$, note that $\angle HA_2I=\angle HB_2I=\angle HC_2I=90$ therefore $IA_2B_2HC_2$ is cyclic. Now $\angle B_2A_2C_2=180-\angle BIC=90-\frac{\angle BAC}{2}=\angle B_1A_1C_1$, now by the same way we get $\angle A_2B_2C_2=\angle A_1B_1C_1$ and $\angle A_2C_2B_2=\angle A_1C_1B_1$ therefore $\triangle A_1B_1C_1 \sim \triangle A_2B_2C_2$ as desired thus we are done .

The points $A_2$, $B_2$, and $C_2$ lie on the circle with diameter $IH$. Notice that $AI$, $BI$, and $CI$ are perpendicular to the sides of $A_1B_1C_1$. Thus $$\measuredangle (A_2B_2,B_2C_2)=\measuredangle(A_2I,C_2I)=\measuredangle(AI,CI)=\measuredangle(B_1C_1,A_1B_1)$$

We solve this by complex bashing. Take the incircle of $\triangle ABC$ as the unit circle, and let $d,e,f$ be the coordinates of $A_1,B_1,C_1$ respectively so that we don't have to write a ton of subscripts. Let $O$ be the circumcenter of $\triangle ABC$. Then $$|d|=|e|=|f|=1$$$$a=\frac{2ef}{e+f}$$$$b=\frac{2df}{d+f}$$$$c=\frac{2de}{d+e}$$$$o=\frac{2def(d+e+f)}{(d+e)(d+f)(e+f)}$$$$h=a+b+c-2o = \frac{2(d^2e^2+d^2ef+d^2f^2+de^2f+def^2+e^2f^2)}{(d+e)(d+f)(e+f)}$$$$e_a = \frac{a+h}2 = \frac{d^2e^2+2d^2ef+d^2f^2+2de^2f+2def^2+2e^2f^2}{(d+e)(d+f)(e+f)}$$$$e_b = \frac{b+h}2 = \frac{d^2e^2+2d^2ef+2d^2f^2+2de^2f+2def^2+e^2f^2}{(d+e)(d+f)(e+f)}$$$$e_c = \frac{c+h}2 = \frac{2d^2e^2+2d^2ef+d^2f^2+2de^2f+2def^2+e^2f^2}{(d+e)(d+f)(e+f)}$$Now $A_2$ is the other point (besides $A$) on the perpendicular bisector of $\overline{B_1C_1}$ that is the same distance as $A$ from $E_A$. So it is the reflection of $A$ over the projection of $E_A$ onto this perpendicular bisector. Thus \begin{align*} a_2 &= 2\left[\frac12\left(\sqrt{ef} - \sqrt{ef} + e_a - \sqrt{ef}(-\sqrt{ef})\overline{e_a}\right)\right] - a \\ &= e_a + ef\overline{e_a} - a \\ &= \frac{d^2e^2+2d^2ef+d^2f^2+2de^2f+2def^2+2e^2f^2}{(d+e)(d+f)(e+f)} + \frac{ef(2d^2+2de+2df+e^2+2ef+f^2)}{(d+e)(d+f)(e+f)} - \frac{2ef}{e+f} \\ &= \frac{d^2e^2+2d^2ef+d^2f^2+2de^2f+2def^2+e^3f+2e^2f^2+ef^3}{(d+e)(d+f)(e+f)} \\ &= \frac{d^2e+d^2f+2def+e^2f+ef^2}{(d+e)(d+f)} \end{align*}Similarly, \begin{align*} b_2 &= \frac{de^2+e^2f+2def+d^2f+df^2}{(d+e)(e+f)} \\ c_2 &= \frac{df^2+ef^2+2def+d^2e+de^2}{(d+f)(e+f)} \end{align*}Then we have the vector \begin{align*} a_2-b_2 &= \frac{d^2e+d^2f+2def+e^2f+ef^2}{(d+e)(d+f)} - \frac{de^2+e^2f+2def+d^2f+df^2}{(d+e)(e+f)} \\ &= \frac{(e+f)(d^2e+d^2f+2def+e^2f+ef^2) - (d+f)(de^2+e^2f+2def+d^2f+df^2)}{(d+e)(d+f)(e+f)} \\ &= \frac{-d^3f-d^2f^2-df^3+e^3f+e^2f^2+ef^3}{(d+e)(d+f)(e+f)} \\ &= -\frac{f(d-e)(d^2+de+e^2+df+ef+f^2)}{(d+e)(d+f)(e+f)} \end{align*}Similarly, $$a_2-c_2 = -\frac{e(d-f)(d^2+de+e^2+df+ef+f^2)}{(d+e)(d+f)(e+f)}$$and so $$\frac{a_2-b_2}{a_2-c_2} = \frac{f(d-e)}{e(d-f)} = \overline{\left(\frac{d-e}{d-f}\right)}$$Thus $\triangle A_1B_1C_1 \overset{-}\sim \triangle A_2B_2C_2$. $\blacksquare$