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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