问题:在Rt△ABC中,∠A=90°,AB=AC,E、F两点在边BC上,且∠EAF=45°,则EF~2=BF~2+EC~2.一、问题的证明证明1如图1,将△AEC绕点A旋转90°得△ABE′,连结E′F.则∠E′BF=∠E′BA+∠ABF=∠C+∠ABF=90°,BE′=EC.在△E′BF中,由勾股定理得E′F~2=BE′~2+BF~2=EC~2+BF~2,在△E′AF和△EAF中,∵∠E′AF=∠E′AB+∠BAF=∠EAC+∠BAF=90°-∠FAE=45°,AE′=AE,AF公用,∴△E′AF≌△EAF.∴EF=E′F,∴EF~2=E′F~2=EC~2+BF~2. Problem: In Rt △ ABC, ~A = 90 °, AB = AC, E, F two points on the edge BC, and ∠EAF = 45 °, then EF ~ 2 = BF ~ 2 + EC ~ PROOF OF PROBLEMS PROOF 1 As shown in Fig. 1, △ AEC is rotated by 90 ° around point A to obtain △ ABE ’, connecting E’F. Then E’BF = ∠E’BA + ∠ABF = ∠C + ∠ABF = 90 °, BE ’= EC. In △ E’BF, by the Pythagorean theorem E’F ~ 2 = BE’ ~ 2 + BF ~ 2 = EC ~ 2 + BF ~ 2. In △ E’AF and △ EAF, ∠E’AF = ∠E’AB + ∠BAF = ∠EAC + ∠BAF = 90 ° -∠FAE = 45 °, AE ’= AE, AF common, ∴ΔE’AF≌ΔEAF.∴EF = E’F, ∴EF ~ 2 = E’F ~ 2 = EC ~ 2 + BF ~ 2.