文献[1]中,有以下6个命题:命题1如图1,在△ABC中,AB=AC,∠BAC=90°,点D1、D2是边AC上两点,且AD1=CD2,AE1⊥BD1交BC于点E1,AE2⊥BD2交BC于点E2.则CE2BE2+CE1BE1=1.图1图2命题2如图2,在△ABC中,AB=AC,∠BAC=90°,点D1在AC上,点D2在AC的延长线上,且AD1=CD2,AE1⊥BD1交BC于点E1,AE2⊥BD2交BC于点E2,则CE2BE2-CE1BE1=1.命题3如图3,在△ABC中,AB=AC,∠BAC=90°,点D在CA的延长线上, In [1], there are 6 propositions as follows: Proposition 1 As shown in Fig.1, in △ ABC, AB = AC and ∠BAC = 90 °. Points D1 and D2 are two points on AC and AD1 = CD2 and AE1 ⊥ BD1 intersection BC at point E1, AE2 ⊥ BD2 intersection BC at point E2. Then CE2BE2 + CE1BE1 = 1. Figure 1 Figure 2 Proposition 2 as shown in Figure 2, △ ABC, AB = AC, ∠ BAC = 90 °, D1 is on AC, point D2 is on the extension of AC, and AD1 = CD2, AE1⊥BD1 at BC, point E1, and AE2⊥BD2 at BC at point E2, then CE2BE2-CE1BE1 = 1. Proposition 3 As shown in Figure 3, In ABC, AB = AC, BAC = 90 °, point D is on the extension of CA,