As shown in the figure, in △ ABC, the bisectors ad, be and CF of the three inner angles intersect at point I, IH ⊥ BC, to verify, ∠ bid = ∠ HIC

It is proved that: ad bisection ∠ BAC ∠ Bai ＝ BAC / 2 ∵ be bisection ∈ ABC ≠ CBI ＝ ABC / 2 ∵ CF bisection ∈ ACB ≠ BCI ＝ ACB / 2 ∵ bid = ∠ Bai + ∠ Abe = (∠ BAC + ∠ ABC) / 2 ∵ BAC + ∠ ABC = 180 - ∠ ACB ∵ bid = (180 - ∠ ACB) / 2 = 90 - ∠ ACB / 2 = 90 - ∠ B

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