Correct answer: Rd4+1.Qg4+? 1...Kb3/[б]//Ka3[d] 2.Rd3#[A] 1...Kc5[c] 2.Qd4#[Б] 1...Ka5[e] 2.Re5#[C] but 1...Kc3[a]! 1.Qd3[D]? zz 1...Kc5[c] 2.Q