TITLE: Prove: In a Triangle, $II_1 = a\cdot \sec \frac{A}{2}$
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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.