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如图1,在△ABC中,∠ABC,∠ACB的角平分线交O,则∠BOC=90°+ 1 2 ∠A= 1 2 ×180°+ 1 2 ∠A. 如图2,在△ABC中,∠ABC,∠ACB的两条三等分角线分别对应交于O 1 ,O 2 ,则∠BO 1 C= 2 3 ×180°+ 1 3 ∠A,∠BO 2 C= 1 3 ×180°+ 2 3 ∠A. 根据以上阅读理解,你能猜想(n等分时,内部有n-1个点)(用n的代数式表示)∠BO n-1 C=( ) A. 2 n ×180°+ 1 n ∠A B. 1 n ×180°+ 2 n ∠A C. n n-1 ×180°+ 1 n-1 ∠A D. 1 n ×180°+ n-1 n ∠A