Question

如图,在三角形ABC中,角ACB等于90度,CD垂直AB,CE=CF,证明BE是角ABC的平分线

Answer 1

∵CE=CF

∴镇则∠CEF=∠CFE=∠DFB

即∠CEB=∠DFB

∵CD⊥AB，那么∠橘告BDC=∠ACB=90°

即∠ECB=∠BDF=90°

∴∠CBE=90°-∠CEB

∠DBF=90°-∠DFB

∴∠圆旅明CBE=∠DBF=∠ABE

∴BE平分∠ABC

Answer 2

如图所示
∵ AC⊥BC,CD⊥AB,
∴顷庆 ∠ABC+∠CAB=∠ABC+∠DCB=90度,
∴ ∠CAB=∠DCB,-------(1)
又 ∠CEF=∠伏码CAB+∠ABE,∠CFE=∠EBC+∠DCB,
∵ CE=CF,
∴ ∠CEF=∠CFE,
即 ∠CAB+∠ABE=∠EBC+∠DCB-----(2)
由上面(1)知,∠缺乎哪CAB=∠DCB,
代入(2),即得
∠ABE=∠EBC
从而,BE是角ABC的角平分线.