Distance between incentre and excentre of a triangle I am unable to get anywhere regarding the distance between the incentre and an excentre of $\triangle ABC$. I do know that it is generalised as: 

$$IJ_c=c\sec\left(\frac C2\right)$$
  For the incenter $I$ and excenter $J_c$, where side $c$ is opposite $\angle C$.

But how can I prove this? Kindly assist.
 A: Consider this diagram:

Consider that the distance between $I$ and $E_c$ is simply:
$$d=E_cC-IC\tag{1}$$
And we have:
$$\cos \left(\frac C2\right)=\frac{KC}{E_cC}\implies E_cC=\frac{KC}{\cos(C/2)}\\
\cos\left(\frac C2\right)=\frac{T_bC}{IC}\implies IC=\frac{T_bC}{\cos(C/2)}$$
However, we know that $T_bC=\frac{b+a-c}2$. This makes $IC=\frac{b+a-c}{2\cos (C/2)}$, and from $(1)$, we get:
$$d=\frac{KC}{\cos(C/2)}-\frac{b+a-c}{2\cos(C/2)}\tag{2}$$
However, looking at the diagram, we know that $KC=b+E_cA\sin(A/2)$, because $KC=b+AK$, and $AK$ is easily derivable using law of sines for right triangles. Therefore, $(2)$ becomes:
$$d=\frac{b+E_cA \sin(A/2)}{\cos(C/2)}-\frac{b+a-c}{2\cos(C/2)}\\
\implies d=\frac{b+E_cA\sin(A/2)-\frac{b+a-c}{2}}{\cos (C/2)}$$
However, from the original statement $d=\frac{c}{\cos (C/2)}$, what we only need to show is that:
$$b+E_c A\sin(A/2)-\frac{b+a-c}2=c\tag{3}$$

Polishing $(3)$ we get:
$$b-a+2E_cA\sin(A/2)=c$$
This implies that in order to get $c$, we need:
$$E_cA\sin (A/2)=\frac{a+c-b}{2}\tag{4}$$

From the diagram, consider that $\sin (A/2)$, can be written as:
$$\frac{\sin(C/2)}{IA}=\frac{\sin (A/2)}{IC}\implies \sin (A/2)=\frac{IC \sin (C/2)}{IA}\tag{5}$$
Now, see that $IT_b$ is essentially the inradius, which is equal to:
$$r=IT_b=IC \sin (C/2)$$
Knowing that the inradius $r$ can be written as (which is easily provable):
$$r=\frac{ab\sin C}{a+b+c}$$
Using the value of the updated $(5)$, and substituting it in $(4)$, we get:
$$E_cA\left(\frac{ab\sin C}{(a+b+c)IA}\right)=\frac{a+c-b}2$$
After some, rearrangement, we get:
$$\frac{E_cA}{IA}=\frac{(a+b+c)(a-b+c)}{2ab\sin C}\tag{6}$$
Now, consider the right triangle $E_cAI$ and after some angle-chasing, you will get that $\angle I=\frac{a+c}2$. Notice that simply, $\tan \angle I=\frac{E_cA}{IA}$. Using this information, we get:
$$\tan\left(\frac{A+C}2\right)=\frac{(a+b+c)(a-b+c)}{2ab\sin C} \tag{7}$$
Expanding the RHS, we get:
$$\tan\left(\frac{A+C}2\right)=\frac{a^2-b^2+c^2+2ac}{2ab\sin C} \tag{8}$$
Using the cosine law, we can show that $2ac\cos B=a^2+c^2-b^2$, substituting this in $(8)$ and we get:
$$\tan\left(\frac{A+C}2\right)=\frac{2ac\cos B+2ac}{2ab\sin C}=\frac{c(1+\cos B) }{b\sin C}$$
Now replacing $B=\pi - A-C$, we get:
$$\tan\left(\frac{A+C}2\right)=\frac{c(1-\cos{(A+C)})}{b\sin C}$$
Using the half-angle formula for $\tan$, we get:
$$\frac{\sin (A+C)}{1+\cos(A+C)}=\frac{c(1-\cos{(A+C)})}{b\sin C}\\
b\sin(A+C)\sin C=c(1-\cos^2(A+C))\\
b\sin (A+C)\sin C=c(\sin^2(A+C))\\
\sin(A+C)=\frac{b\sin C}{c}$$
However, using the sine law, we can write $b$ as $b=\frac{c \sin B}{\sin C}$, thus we get:
$$\sin(A+C)=\sin B$$
Which is always true since:
$$B=\pi-A-C\implies \sin(\pi-A-C)=\sin(A+C)\\
Q.E.D.$$

This proof was very long winded, however. There are many ways to write the distance between the excenters and incenters. For example, you have:
$$d=\frac{r_x-r}{\sin(C/2)}$$
For the exradius $r_x$ and in radius $r$. Also, you have $d$:
$$d=\frac{r}{\sin(\frac A2)\cos(\frac{A+C}2)}$$
