Discussed before. AD is radical axis of the reflections of the B-excircle and C-excircle on the midpoints of AC and AB. http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=35317 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=142767 (b)

Doable without the excircles. Let $A_1$ be the excenter opposite $A$, $B_2C_2\cap BC\equiv X$, $B_1C_1\cap BC\equiv E$. Note $ABC$ is the orthic triangle of $A_1B_1C_1$, and $AC_1BC_2, AB_1CB_2$ are both parallelograms. Thus $\triangle BXC_2\sim \triangle CXB_2\implies \frac{CX}{CB_2}=\frac{CX+BX}{CB_2+BC_2}=\frac{BC}{B_1C_1}=\frac{A_1D}{A_1A}$(the last equality due to the two lengths being the two altitudes of $\triangle A_1BC\sim \triangle A_1B_1C_1$). Note the cyclic quadrilateral $EADA_1$(with diameter $EA_1$) $\implies \angle AA_1D=\angle AEB=\angle BCB_2\implies \triangle CXB_2 \sim \triangle A_1DA$. Since $CX\perp A_1D, AA_1\perp B_2C$, the two triangles are a $90^{\circ}$ rotation of each other $\implies AD\perp B_2X\equiv B_2C_2$.

Let $M$ and $N$ be the midpoints of $BB_1$ and $CC_1$. Then using vectors, we have \[b_2-c_2=(a+c-b_1)-(a+b-c_1)=(c+c_1)-(b+b_1)=2n-2m,\]and so it follows that $\overline{B_2C_2}$ is parallel to $\overline{MN}$. Let $A_1$ be the $A$-excenter and let $H$ be the orthocenter of $A_1BC$. Since $\overline{MN}$ is the Gauss line of quadrilateral $BCB_1C_1$, it is perpendicular to its Aubert line which is $\overline{IH}$. This passes through the midpoint of $BC$ by the orthocenter lemma, say $P$, and it is known that $IP\parallel AD$, so we are done.

Apply barycentric coordinates with respect to $\triangle ABC$. Then $B_2=A+C-B_1=\left(1-\frac{a}{a-b+c},\frac b{a-b+c},1-\frac c{a-b+c}\right)=\left(\frac{c-b}{a-b+c},\frac b{a-b+c},\frac{a-b}{a-b+c}\right)$. Similarly, $C_2=\left(\frac{b-c}{a+b-c},\frac{a-c}{a+b-c},\frac{c}{a+b-c}\right)$. Now $\overrightarrow{AD}=\left(-2a,a-b+c,a+b-c\right)$ and $\overrightarrow{B_2C_2}=\left(2a(b-c),a^2-2ab+2bc-c^2-b^2,c^2-a^2+2ac+b^2-2bc\right)$. Note that $a^2[(a-b+c)(c^2-a^2+2ac+b^2-2bc)+(a+b-c)(a^2-2ab+2bc-c^2-b^2)]\\ +b^2[-2a(c^2-a^2+2ac+b^2-2bc)+2a(b-c)(a+b-c)]\\ +c^2[-2a(a^2-2ab+2bc-c^2-b^2)+2a(b-c)(a-b+c)]\\ =a^2[ac^2-a^3+2a^2c+ab^2-2abc-bc^2+a^2b-2abc-b^3+2b^2c+c^3-a^2c+2ac^2+b^2c-2bc^2+a^3-2a^2b+2abc-ac^2-ab^2+a^2b-2ab^2+2b^2c-bc^2-b^3-a^2c+2abc-2bc^2+c^3+b^2c]\\ +b^2[-2ac^2+2a^3-4a^2c-2ab^2+4abc+2a^2b+2ab^2-2abc-2a^2c-2abc+2ac^2]\\ +c^2[-2a^3+4a^2b-4abc+2ac^2+2ab^2+2a^2b-2ab^2+2abc-2a^2c+2abc-2ac^2]\\ =a^2[4b^2c+2c^3+2ac^2+2b^2c-2bc^2-2ab^2-4bc^2-2b^3]\\ +b^2[2a^3-4a^2c+2a^2b-2a^2c]+c^2[-2a^3+4a^2b+2a^2b-2a^2c]=\\ 4a^2b^2c+2a^2c^3+2a^3c^2+2a^2b^2c-2a^2bc^2-2a^3b^2-4a^2bc^2-2a^2b^3+2a^3b^2-4a^2b^2c+2a^2b^3-2a^2b^2c-2a^3c^2+4a^2bc^2+2a^2bc^2-2a^2c^3=0.$

Generalization (Reflections are orthocenters): https://artofproblemsolving.com/community/c6h2374067_ah___h_bh_c_wanted_2_orthocenters__projection_circumcircle_related