Second, as shown in Figure 2, ∠ ABC = 54, ∠ BAC = 70, BE and AD are the heights of △ABC, and H is the intersection of BE and AD, so find the degree of ∠BHA.

Question 1: Take a little f from BC, make BF=AB, connect e f, and it is easy to prove △ BFE △ BAE, ∴∠ AEB = ∠ FEB...① Again? ∠ 1+∠ 2+∠ 3+∠ 4 = 180, while ∠ 1=∠2, ∠3=∠4, ∠.