Question

如图,在三角形ABC中,∠C=90o,外角∠EAB,∠ABF的角平分线AD,BD相交于点D

Answer 1

∠EAB+∠BAC+∠ABF+∠ABC=2*180

(∠EAB+∠ABF)+(∠BAC+∠ABC)=360

(2∠DAB+2∠DBA)+90=360

2(∠DAB+∠DBA)=270

2(180-∠D)=270

2∠D=90

∠D=45