As shown in the figure, in △ ABC, ab = AC, e is on AC, and the extension line of ad = AE, de intersects BC at point F

As shown in the figure, in △ ABC, ab = AC, e is on AC, and the extension line of ad = AE, de intersects BC at point F

It is proved that: as shown in the figure, am ⊥ BC is made in M through a, ∵ AB = AC, ∵ BAC = 2 ⊥ BAM, ∵ ad = AE, ∵ d = ∠ AED, ∵ BAC = 2 ⊥ BAM = 2 ⊥ D, ∵ BAM = D, ∵ DF ∥ am, ≁ am ⊥ BC, ≁ DF ⊥ BC