如图,已知E是三角形ABC的内心(即角平分线交点)角BAC的平分线交BC于点F,且与三角形ABC的外接圆相交于点D

1,
∠BAE=∠CAD ∠ABE=∠EBC
∠DEB=∠BAE+∠ABE=∠CAD+∠EBC ∠CAD=∠CBD
∠DEB=∠CBD+∠EBC=∠DBE
故∠DBE=∠DEB
2,连接CD
∠BAD=∠CAD 弧BD=弧CD ∠BCD=∠DCB
BD=CD ∠DBE=∠DEB BD=DE
DE=CD
∠BCD=∠DCB ∠CBD=∠CAD
∠BCD=∠CAD
即∠FCD=∠CAF ∠CDF=∠ADC
△FCD∽△CAD
DF:CD=CD:AD
CD^2=DF*AD
AD=8 DF:AF=1:3 DF:AD=1:4
DF=8*1/4=2
CD^2=2*8=16
CD=4
DE=CD=4