A circle, two tangents and a triangle - finding incircle center of triangle Let be $K$ a circle. Two tangents  touch the the circle at $C$ and $D$ cross at a point $E$.
So there is a triangle $CDE$.
How do I show that the incircle center (where the bisectors of the triangle cross) is ON the circle $K$ ??
It's clear if it is drawn out..but I have  no idea how to show that.

 A: Let's consider the characterization of the incenter as the intersection of the angle bisectors of the triangle. We claim the desired incenter is in fact the midpoint of minor arc $CD$. It suffices to check that two of the angle bisectors contain this point, as it gives us the third for free.

*

*Note that since the angle bisector at $C$ splits the angle at $C$ in half, the two arcs subtended by those angles must be equal, so the angle bisector intersects the midpoint of the arc.

*The angle bisector at $D$ follows the same reasoning as that at $C$.

Therefore the incenter is the midpoint of minor arc $CD$, as desired.
A: There is symmetry in the diagram  due to equal tangent lengths forming an isosceles triangle.
$$ (\sin Q= s\; ; \cos Q= c\;) \text{ for short} $$
Computing length $(x,H)$ by two different methods. to locate the common point of concurrency.

*

*Via Bisector length using trig formula $ B=\dfrac{2 bc}{b+c} \cos A/2$ at vertex  with angle A

$$ B= \frac{2 c c/s}{c+c/s} cos P= \frac{2 \;c \;c/s}{c+c/s} \frac{c}{B} $$
$$ \text{cross-multiply } B^2=\frac{2c^2}{1+s};\; B=\sqrt{\frac{2 c^2}{1+s}}$$
Using Pythagoras thm
$$ x=\sqrt{B^2-c^2}=\sqrt{\frac{2c^2}{1+s}-c^2}=(1-s) \text{ on simplification}\tag 1$$


*Via in-radius formula for triangle enclosed between tangents and chord. $ r= \dfrac{\Delta}{s}$ involves semi-perimeter and Area

$$ \text{2 semi-perimeter}= 2c+2 c/s,\;\text{semi-perimeter} = c(1+1/s)$$
In-radius
$$ \Delta = c\; c^2/s=c^3/s\;$$
$$r=\frac{c^3/s}{c/s(1+s)}= (1-s) \tag 2 $$
Since
$$x=r=H = 1- \sin Q,$$
so the bisectors B must meet on the middle point of arc of circle K at I.

