Let s = (a+b+c)/2 the semiperimeter

r radius of incircle, centered in I, R radius of circumcircle, centered in O.

r

points I,J_{a},B,C are on a same circle

Let K the midpoint of arc BC in circumcircle. It lies on the perpendicular bisector hence KB = KC.
It lies on the angle bisector as inscribed angles KBC = KAC = A/2.
We also have CBI = B/2, hence angle KBI = (A+B)/2

As BKA = BCA = C, this results into angle BIK = (A+B)/2, hence triangle BKI is isosceles and KB=KI.

Hence K is the center of circle BICJ_{a}

Midpoint of IJ_{a} is on the circiumcircle

BA' = BC' = s - b, CA' = CB' = s - c, AB' = AC' = s - a

and AC" = AB" = s, CA" = CB" = s - b, BA" = BC" = s - c

We deduce at once A'A" and BC have same midpoint M

As A' and C' are symmetric through BI, A'C' is parallel to BJ_{a} and
Triangles A'B'C' and J_{a}J_{b}J_{b} are similar

Also mention the direct consequence of Ceva theorem :

Lines from vertices to intouch points are concurrent (Gergonne point).

Line from vertices to inner touch points with excircles are concurrent (Nagel point).

Similarily for excircle A = 1/2 (b + c - a).r

Putting together this with Heron relation A =

r =^{ }√(s-a)(s-b)(s-c)/s
and for excircle :
^{ }r_{a} = √s(s-b)(s-c)/(s-a)

We also get the relations :

A = √(r.r_{a}.r_{b}.r_{c})
and 1/r = 1/r_{a} + 1/r_{b} + 1/r_{c}

Let Q the diametraly opposite point to K on the circumcircle. Inscribed angles BQK = BAK, triangle BQK is right angled in B, hence similar to triangle C'AI, hence BK/C'I = KQ/IA, that is BK.IA = 2Rr

As IK = BK (K center of circle BICJ

OI² - R² = -2Rr that is : OI² = R² - 2Rr

A similar calculation with excenter gives OJ_{a}² = R² + 2Rr_{a}

BC is common tangent to incircle and excircle.

The second common tangent δ is by reflection through center line IJ, hence goes through K. This line is antiparallel to side BC with respect to the two other sides.

Hence it is parallel to the tangent in A to circumcircle.

In triangle BKA, angle bisectors from B intersect side AK in I and J,
hence (AKIJ) is an harmonic range : IK/IA = JK/JA = BK/BA

Perpendicular projection on line BC gives then an harmonic range (HKA'A")

As M is midpoint of BC and A'A", MA'² = MA"² = MK.MH

MA'² is also the power of M to the incircle, as well as MA"² is the power to the excircle.

Let's consider an inversion with pole M and power MA'² = MA"².
This then let's unchanged the incircle and the excircle. It transforms δ
into a circle going at pole M and through the inverse of K, that is H (as MK.MH = MA'²),
and still tangent to incircle and excircle.

The tangent in M to that circle is parallel to δ,
hence parallel to the tangent in A to circumcircle.

Also consider the Euler circle of triangle ABC.
This circle goes through M and H.

A dilation of center G (centroid) and ratio -1/2
transforms the circumcircle into the Euler circle, and point A into M.

The tangent in M to Euler circle is then parallel to the tangent in A to circumcircle (dilation !)

The two considered circles : Euler circle and inverse of δ both go through M and H,
and have the same tangent in M, hence are the same circle.
Hence the Feuerbach theorem :

The Euler circle touches the incircle and the excircles

The contact points are named "Feuerbach points".