Let be an acute triangle with circumcircle . Let be a tangent line to , and let and be the lines obtained by reflecting in the lines , and , respectively. Show that the circumcircle of the triangle determined by the lines and is tangent to the circle .

Proof:

Let be tangent to at and let and and Let and and WLOG assume that the order of points on line is (to avoid configuration issues). Let be the incenter of

Lemma 1:

Proof: Consider It is clear that line is the internal angle bisector of and it is also clear that line is the internal angle bisector of Therefore is the incenter of this triangle; hence, line bisects Similarly line bisects so we have the desired result.

Lemma 2:

Proof: Note that $ \angle{AC_2B_2} + \angle{A_B_2C_2} = 180 - \angle{BAC} $ which implies the desired result.

Now Let be the center of and let and be the reflections of about and respectively. Clearly and are on

Lemma 3:

Proof: Note that which immediately implies the desired result.

Letting be the reflection of about we have that and as well. Therefore there exists a homothety taking to It suffices to show that the center of this homothety lies on , because then the homothety will take to the circumcircle of and so will be the tangency point between these two circles.

Now let be the reflection of about Cleary Since we have that and similarly that Therefore Now let be the second intersection of line and and let It suffices to show that By Pascal's Theorem on cyclic hexagon we have that are collinear so lies on line which implies that so is the desired tangency point and we are done.