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Let ”ABC” be a triangle, ”I” its in-centre; $A_1, B_1, C_1$ be the reﬂections of I in BC, CA, AB respectively. Suppose the circumcircle of triangle $A_1 B_1 C_1$ passes through A. Prove that $B_1, C_1, I, I_1$ are concyclic, where $I_1$ is the incentre of triangle $A_1 B_1 C_1$.

Basic Sketch:

Hint 1: I is indeed the circumcenter of $A A_1 B_1 C_1$ with circum radius = 2r. $B_1 C_1$ is the radical axis of the two circles concerned hence the other center has to lie of IA (since IA is perpendicular to radical axis and I is one of the centers hence IA is the line joining the centers).

Hint 2: We prove that A is the center of the circle $B_1, C_1, I, I_1$ . Using cosine we show that $\Delta AIH$ is a 30-60-90 triangle as IH = r and IA = 2r. This implies $\angle B_1 I C_1$ is $120^o$ . Hence it is sufficient to show $\angle B_1 I_1 C_1$ is $120^o$ . A simple angle chasing in triangle $A_1 I_1 C_1$ and $B_1 I_1 C_1$ completes the proof (we observe that $\angle A_1 C_1 B_1 = \frac {A+B} {2}$ and similarly with the other angles).

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