Question

BD是角ABC的平分线,角EAB是外角求证角BDE等于2分之一的角C+角BAE

Answer 1

设∠ABD=∠CBD=∠1

则∠EAB=2∠1+∠C①

∠BDE=∠1+∠C

2∠BDE=2∠1+2∠②

②－①

2∠BDE=∠C+∠BDE

∠BEE=1/2(∠C+∠BDE)