$ CI_c$ cuts $ AB$ at $ Z$ and $ I \in CI_c$ is incenter of $ \triangle ABC.$ Cross ratio $ (C, Z, I, I_c) = - 1$ is harmonic and $ T$ is midpoint of $ II_c$ $ \Longrightarrow$ $ \frac {CT}{TI_c} = \frac {CI}{IZ} = \frac {a + b}{c}.$ $ AI_c$ cuts $ BC$ at $ X$ and $ AI_c = \frac {c - b}{a + b - c} \cdot AX = \sqrt {bc \cdot \frac {c + a - b}{a + b - c}}.$ $ BA = BD$ and $ BI_c$ bisects $ \angle ABD$ $ \Longrightarrow$ $ ABDI_c$ is a kite, $ \angle CDI_c = \angle BDI_c = \angle BAI_c = \frac {_1}{^2}(\angle B + \angle C).$ $ \angle TDI_c = \frac {_1}{^4} (\angle B + \angle C) \Longleftrightarrow$ $ DT$ bisects $ \angle CDI_c \Longleftrightarrow$ $ \frac {CT}{TI_c} = \frac {DC}{DI_c} = \frac {DB + BC}{AI_c} \Longleftrightarrow$ $ \frac {(a + b)^2}{c^2} = \frac {(c + a)^2 (a + b - c)}{bc (c + a - b)} \Longleftrightarrow$ $ (c - b)(a + b + c)(a^2 - b^2 + bc - c^2) = 0 \Longleftrightarrow$ $ b = c$ or ($ a^2 = b^2 - bc + c^2 \Longleftrightarrow \angle A = 60^\circ$).

I have an easy solution which doesn't need calculation. $ \angle ADC = \angle AI_{c}C = \frac {\angle B}{2}$ Thus $ CDI_{c}A$ is a cyclic so $ \angle CDI_{c} = \frac {\angle B + \angle C}{2}$ but $ BD\ne DI_{c}$ and $ BT = TI_{c}$ which means $ DBTI_{c}$ is a cyclic quadrilateral . now $ \angle BDI_{c} + \angle BTI_{c} = 180$ and we get : $ \angle A = 60^\circ$

I have to remember this idea.

it is easy problem. $ BC=BD,BI_c=BI_c,\angle I_cBA=\angle I_cBD$ so $ \triangle AI_cB \cong \triangle DI_cB$. so $ \angle BDI_c=\angle BAI_c,\angle AI_cB=\angle BI_cD$. it is easy to measure this angle: $ \angle BDI_c=\angle BAI_c=\frac{\angle B +\angle C}{2}$ $ \angle AI_cB=\angle BI_cD=\frac{\angle A +\angle B}{2},\angle TI_cB=\frac {\angle A}{2},\angle TBC=\angle I_cCB=\frac{\angle C}{2}$ and $ AT=BT=I_cT$(1). because of $ \angle TDI_c=\frac {\angle B+\angle C}{4}$ so $ \angle TDI_c=\angle TDC =\frac {\angle B+\angle C}{4}$ so $ DT$ bisector of $ \angle I_cDC$ so $ \frac {TI_c}{TC}=\frac {DI_c}{DC}$(2). with (1): $ \frac {TI_c}{CT}=\frac {TB}{TC}=\frac {\sin \frac {C}{2}}{\sin \frac {2B+C}{2}} , \frac {DI_c}{DC}=\frac {\sin \frac {C}{2}}{\sin \frac {2A+B}{2}}$(3). with (2),(3): $ \frac {\sin \frac {C}{2}}{\sin \frac {2B+C}{2}} = \frac {\sin \frac {C}{2}}{\sin \frac {2A+B}{2}}$ so ${{ \sin \frac {2B+C}{2}}=\sin \frac {2A+B}{2}}$ so ${{ \frac {2B+C}{2}}=\frac {2A+B}{2}}$ or ${{ \frac {2B+C}{2}}+\frac {2A+B}{2}}=180$ so $ \angle A=60^{\circ}$ or $ \angle B=\angle C$ so $ \angle A=60^{\circ}$(because of $ AB \neq AC$)

Let $X$ and $Y$ be the points of contact of the excircle with $AB$ and $BC$.Then note that $BX=BY=s-a$ so $BX=YD$.Thus $\triangle{AXI_c} \cong \triangle{DYI_c}$ so $\angle{IDY}=90-\frac{A}{2}$.It is given that $\angle{TDI_c}=\frac{B+C}{4}=45-\frac{A}{4}$ so $DT$ bisects $\angle{CDI_c}$. Now note that $DI_c=\frac{r_c}{cos\frac{A}{2}}=\frac{s-b}{tan\frac{A}{2}cos\frac{A}{2}}=\frac{s-b}{sin\frac{A}{2}}$ Also we have $CD=a+c$ Thus $\frac{I_cD}{CD}=\frac{s-b}{(a+c)sin\frac{A}{2}}=\frac{a-b+c}{2(a+c)sin\frac{A}{2}}=\frac{sinA+sinC-sinB}{2(sinA+sinC)sin\frac{A}{2}}=\frac{cos\frac{A-C}{2}-cos\frac{A+C}{2}}{2sin\frac{A}{2}cos\frac{A-C}{2}}=\frac{sin\frac{C}{2}}{cos\frac{A-C}{2}}$ Easy angle chasing gives $I_cT=BT$.Thus $\frac{I_cT}{CT}=\frac{BT}{CT}=\frac{sin\frac{C}{2}}{sin(B+\frac{C}{2})}$ Now ${\frac{DI_c}{CD}=\frac{I_cT}{CT} \implies sin(90+\frac{C}{2}-\frac{A}{2})=cos\frac{A-C}{2}=sin(B+\frac{C}{2})}=sin(A+\frac{C}{2})$ So $90+\frac{C}{2}-\frac{A}{2}=A+\frac{C}{2} \implies \frac{3A}{2}=60 \implies \angle{A}=60^{\circ}$ or $90-\frac{C}{2}+\frac{A}{2}=A+\frac{C}{2} \implies \frac{A}{2}+C=90^{\circ}=\frac{A}{2}+B \implies B=C$.This is impossible as $AB \neq AC$. From all these we conclude that $\angle{A}=60^{\circ}$