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 $.
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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