Prove: In a Triangle, $II_1 = a\cdot \sec \frac{A}{2}$ Prove that $II_1 = a\cdot \sec \dfrac{A}{2}$.
$I$ is center of incircle, $I_1$ is center of excircle.
What I did is :
Drop $ID \perp AB$, & $I_1F \perp AF$ at $F$ So $ID\parallel I_1F$
$\dfrac{AI}{II_1} = \dfrac{AD}{DF}$
$II_1 = \dfrac{AI \cdot DF}{AD}$
$II_1 = DF \cdot \sec \dfrac{A}{2}$

What should I do further, or provide me another approach.
 A: Let $T$ be the midpoint of $II_1$.  Show that $\angle BTC=\pi-\angle BAC$ (i.e., $ABTC$ is a cyclic quadrilateral).  Prove that $T$ is the center of the circumcircle of the cyclic quadrilateral $IBI_1C$.  The result follows immediately.

A: Let $p$ be the half of perimeter of the triangle (i.e $p = \frac{a+b+c}{2}$). We have $AD = p-a$.
Let compute $BF$. Taking $I_1E \perp BC$ and $I_1G \perp AC$, call that $BE = u$ and $CE = v$, one then has $u+v = a$ and $u + c = v + b$, or $u - v = b-c$. 
Thus, $BF = BE = u = \frac{a+b-c}{2} = p - c$. 
So, one has $DF = AF - AD = AB + BF - AD = c + p-c -(p-a) = a$.
EDIT: to answer your question in comment, you can draw the line perpendicule to $AB$, $AC$, $BC$ from $I$ and $I_1$. So, you have $2AD = 2p - 2a$, then $AD = p-a$.
Similarly, you have $AF = AG$ or $b+u=c+v$.
A: There is an even simpler way in order to deal with the problem.
Now to find $II_1$, you can also use the following approach :
$$II_1=I_1A-IA$$
$$I_1A= r_1/sin(A/2)$$
and similarly:
$$IA= r/sin(A/2)$$
$$\Rightarrow II_1= r_1/sin(A/2)-r/sin(A/2)$$
Replace $r_1$  by $s\cdot (tan(A/2))$ and $r$ by $(s-a)\cdot \tan(A/2)$
Writing the equation again,
$$ II_1=\frac{s\cdot tan(A/2)}{sin(A/2)}-\frac{(s-a)\cdot \tan(A/2)}{sin(A/2)}$$
$$II_1=\frac{\tan(A/2)}{\sin(A/2)}{a}$$
$$II_1=a/cos(A/2)$$
