引例(2013年山东东营市)(1)如图1-1,已知:在△ABC中,∠BAC=90°,AB=AC,直线m经过点A,BD⊥直线m,CE⊥直线m,垂足分别为点D、E.证明:DE=BD+CE.(2)如图1-2,将(1)中的条件改为:在△ABC中,AB=AC,D、A、E三点都在直线m上,并且有∠BDA=∠AEC=∠BAC=α,其中α为任意锐角或钝角.请问结论 DE=BD+CE是否成立?如成立,请你给出证明;若不成立,请说明理由.(3)拓展与应用:如图1-3,D、E是D、A、E三点所在直线m上的两动点(D、A、E三点互不重合),点F为 Example (1) As shown in Figure 1-1, it is known that in △ ABC, ∠ BAC = 90 ° and AB = AC, the straight line m passes through the point A, the straight line BD⊥, the straight line m , And foot drop respectively point D, E. Proof: DE = BD + CE. (2) As shown in Figure 1-2, the conditions in (1) are changed to: AB = AC, D, A, E are three points in the straight line m, and there ∠ BDA = ∠ AEC = ∠ BAC = α, where α is any acute or obtuse angle. Can I conclude whether DE = BD + CE holds? If yes, (3) Development and application: As shown in Figure 1-3, D and E are two moving points on the line m where D, A and E are three points (D, A and E do not coincide with each other at three points) , Point F is