Question

若点P是∠ABC和外角∠ACE的角平分线的交点,求证：∠P=90°-∠A

Answer 1

楼主,你先画个图,再看下面解答

记∠ABC=B,∠ACB=C,∠BAC=A

∠P=∠PCE-∠PBE

=（180°-C）/2 -B/2

=90-（B+C）/2

=90-（180-A）/2=A/2

此题答案应为A/2