Question

下面两大题,我想了很长时间也没想出来,

Answer 1

1、②∠P=∠PCD-∠PBD=1/2×∠ACD-1/2×∠ABC=1/2×（∠ACD-∠ABC）=1/2×∠A

③∠PBC+∠PCB=1/2×∠BCE+1/2×∠CBD=1/2×（∠BCE+∠CBD）=1/2×（180°-∠ACB+180°-∠ABC）=1/2×【360°-（180°-∠A）】=1/2×（180°+∠A）=90°+1/2×∠A

故∠P=180°-（∠PBC+∠PCB）=180°-（90°+1/2×∠A）=90°-1/2×∠A