(1) solution: the bisector of the inner angle of ∵△ABC intersects with point I.

Let ∠ IBA = ∠ IBC = X, ∠ ACI = ∠ ICB = Y?

∠A = 180 -2(X+Y)

X+Y=90- 1/2∠A①

∠BIC= 180 -(X+Y)②

① Substitute into ②

∠BIC= 180 -(90- 1/2∠A)

∠BIC=90？ + 1/2∠A

∴∠BIC=90？ + 1/2∠A