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1. The points B, C, H, I lie on a single circle, then O lies on this circle too.

   For given question,

   We have been given I, O, H be the incenter, circumcenter, and orthocenter of an acute triangle ABC, respectively

   In an acute ΔABC, O is circumcenter. Thus, by applying the inscribed angle theorem to the circle, we have:

   ∠BOA = 2∠BAC.               ………………(1)

   Assuming H is orthocenter; Thus, the angle formed by angles BAC and BC at H are supplementary angles:

   ∠BAC + BHC = 180°.

   BHC = 180° – ∠BAC.           ……………….(2)

   Assuming I is incenter; Thus, the angle formed by BIC is given by:

   ∠BIC = 90° + ∠A/2.            ……………….(3)

   Since points B, C, H, I lie on a single circle and the angles formed in the same side of an arc of a circle are equal, we have:

   BIC = BHC                           ……………….(4)

   Substituting the parameters into equation (4), we have;

   ⇒ 180° – ∠BAC = 90° + ∠A/2

   ⇒ ∠BAC + ∠BAC/2 – 3∠BAC/2 = 180° – 90°

   ⇒ ∠BAC = 90° × 2/3

   ⇒ ∠BAC = 60°.

   From equation (1), we have:

   ⇒ ∠BOA = 2∠BAC.

   ⇒ ∠BOA = 2 × 60

   ⇒ ∠BOA = 120°.

   Hence, this proves that since points B, C, H, I lie on a single circle, then O lies on this circle too.

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