points I,Ja,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 BICJa
Midpoint of IJa is on the circiumcircle
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 BJa and Triangles A'B'C' and JaJbJb 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).
r = √(s-a)(s-b)(s-c)/s and for excircle : ra = √s(s-b)(s-c)/(s-a)
We also get the relations :
A = √(r.ra.rb.rc) and 1/r = 1/ra + 1/rb + 1/rc
OI² - R² = -2Rr that is : OI² = R² - 2Rr
A similar calculation with excenter gives OJa² = R² + 2Rra
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".