Correct answer: Bd71...Qc4[a] 2.Qd6#[A] 1.Bd4? (2.Qc5#) 1...Qc4[a] 2.Qd7#[C] 1...Rc1/[б]/ 2.Bg4#[Б] 1...Rc2[c] 2.Bh3#[D] but 1...Rg7! 1.Ng5
