Question

如图,△ABC的三个内角平分线交于点F,过F作DE⊥AF,垂足为F,交AB,AC于D,E两点,求证∠BDF=∠BFC=∠

Answer 1

易证△AEF≌ADF,∠AEF=∠ADF

∠CEF=∠BDF=90°+∠BAC/2 (外角=不相邻二内角和)

∠BFC=180°-(∠ABC+∠ACB)/2 (角平分线、三角形内角和)

=180°-(180°-∠BAC)/2 (三角形内角和)

=180°-90°+∠BAC/2

=80°+∠BAC/2

∠CEF=∠BDF=90°+∠BAC/2=∠BFC

毕