三角形ABC中,BP CP分别平分角ABC 角ACD 求角P与角A的关系 并证明理由

关系：∠BPC =90°+1/2∠A
证明：在ABC中,∠ABC和∠ACB的平分线相交于点P
所以∠BPC
=180°-（∠PBC+∠PCB)
=180°-（1/2∠ABC+1/2∠ACB)
=180°-1/2(∠ABC+∠ACB)
=180°-1/2(180°-∠A)
=180°-90°+1/2∠A
=90°+1/2∠A