Question

三角形ABC中,延长边BC到D,BP平分角B,CP平分角ACD,求证：角P等于90度减去1/2角A

Answer 1

结论是：∠P=1/2∠A ∠DCP=∠P+∠PBC, ∵BP、CP分别平分∠ABC、∠ACD ∴∠PBC=1/2∠ABC ∠DCP=1/2∠ACD=1/2(∠A+∠ABC) =1/2∠A+1/2∠ABC= ∴∠P+∠PBC=∠1/2∠A+1/2∠ABC ∴∠P=1/2∠A 另外两条内角平分线的交角等于90°+1/2∠A ∠B、∠C的外角平分线交角等于90°-1/2∠A