Question

如图，已知:在三角形ABC中，∠ABC＝3∠C，AD是∠BAC的平分线，BE⊥AD于E.求证AC-AB=2BC

怎么证三角形BFC为等腰。注：不是直角三角形

Answer 1

证明：延长BE交AC于点F
∵BE⊥AE
∴∠AEB=∠AEF=90°
∵∠EAB=∠EAF AE=AE ∠AEB=∠AEF
∴△AEB≌AEF(ASA)
∴BE=FE AB=AF
即BF=2BE
∵∠B=3∠C
∠BAC+∠C+∠ABC=180°
∴∠BAC+∠C+3∠C=90°
∴∠BAC+4∠C=180°
1/2∠BAC+2∠C=90°
∵∠BAE=1/2∠BAE
∠BAE+∠ABE=90°
∴∠ABE=2∠C
又∠ABC=3∠C
∴∠FBC=∠C
∴FC=FB
∴AC-AB=AC-AF=FC=FB=2BE
即AC-AB=2BE

Answer 2

题目打错了吧