everyone! I am confused about what to do here and how to do it. May I have some help?
So, I know that $\overline{FI}$ and $\overline{IE}$ are both radii of the small circle and $\overline{JK}$ and $\overline{JH}$ are both radii of the big circle. I am pretty sure that the outer tangents are equal, but in this picture, they don't look equal...
Now, I can't proceed. Could someone help me?
Question (c) solved... (a) and (b) left...
Thank You!
Extend $AB$, $AC$ to get $\triangle AB_2C_2\sim \triangle ABC$ such that $\bigcirc J$ is the incircle of $\triangle AB_2C_2$.
Then \begin{align} \angle ICB&=\tfrac\gamma2 ,\\ \angle BCC_2&=180^\circ-\gamma ,\\ \angle BCJ&=\tfrac12\angle BCC_2 =90^\circ-\tfrac\gamma2 ,\\ \text{hence, }\quad \angle ICJ&=\angle ICB+\angle BCJ=90^\circ . \end{align}
Edit
Let \begin{align} \frac{|JK|}{|IE|}&= \frac{r_2}{r} =k . \end{align}
Using two expressions for the height $h_2$ of $\triangle AC_2B_2$
\begin{align} h_2&=h\cdot k=\frac{2S}{a}\cdot k ,\\ h_2&=h+2r_2= \frac{2S}a+2r\cdot k , \end{align}
we get $k$ in terms of semiperimeter $\rho$ of $\triangle ABC$: \begin{align} k&=\frac{S}{S-r a} =\frac{\rho}{\rho-a} , \end{align}
thus \begin{align} r_2&=\frac{\rho r}{\rho-a} \end{align}
and we have
\begin{align} |AH|&=r_2\cot\tfrac\alpha2 =r_2\cdot\frac{|AF|}{|IF|} =\frac{\rho r}{\rho-a}\cdot\frac{\rho-a}{r} =\rho . \end{align}
\begin{align} |CD|&=\rho-c ,\\ |BG|&= r_2\cdot \tan\tfrac\beta2 =\frac{\rho r}{\rho-a} =\frac{\rho r}{\rho-a}\cdot \frac{r}{\rho-b} \\ &= \frac{\rho r}{\rho-a}\cdot \frac{r}{\rho-b} \cdot \frac{\rho(\rho-c)}{\rho(\rho-c)} \\ &= \frac{\rho^2 r^2} {\rho(\rho-a)(\rho-b)(\rho-c)} \cdot(\rho-c) =\rho-c =|CD| . \end{align}