As shown in the figure: in △ ABC, ab = AC, take a little E on AC, extend Ba to F, make AF = AE, and connect Fe. At this time, there is a special position relationship between Fe and BC. Can you find out and explain it?

As shown in the figure: in △ ABC, ab = AC, take a little E on AC, extend Ba to F, make AF = AE, and connect Fe. At this time, there is a special position relationship between Fe and BC. Can you find out and explain it?

It is proved that if ad ⊥ BC is used in D, then ∵ BAC = 2 ⊥ bad. ∵ BAC = F + ⊥ AEF, and ∵ AF = AE, ∵ f = AEF, ∵ BAC = 2 ⊥ F, ∵ f = bad, ∵ EF ∥ ad, ≁ EF ⊥ BC