Question

如图,△ABC中,∠C=90°,∠ABC的角平分线与∠BAC的外角平分线相交于E点，求证：∠E=45°

如图,△ABC中,∠C=90°,∠ABC的角平分线与∠BAC的外角平分线相交于E点，求证：∠E=45°

Answer 1

由三角形内角和=180°可知，

∠DAC=∠ABC+∠C；∠C=∠CAB+∠CBA=90°

由已知：∠EAC=1/2∠DAC=1/2∠ABC+1/2∠C；且∠ABE=1/2∠ABC，

∠E=180°-∠EAC-∠CAB-∠ABE

=180°-(1/2∠ABC+1/2∠C)-∠CAB-1/2∠ABC

=180°-∠ABC-∠CAB-1/2∠C

=180°-90°-1/2∠C

=90°-(1/2)·90°

=45°