Question

如图所示,BP,CP分别是∠ABC,∠ACB的平分线,并交于点P,①求角P的度数.②若角A=50°,110°时,角P的值

Answer 1

规律：∠P=90°+1/2∠A.

证明：∵BP,CP分别是∠ABC,∠ACB的平分线,

∴∠PBC=1/2∠ABC,∠PCB=1/∠ACB,

∴∠P=180°-∠PBC-∠PCB (ΔPBC内角和为180°)

=180°-1/2(∠ABC+∠ACB)

=180°-1/2(180°-∠A) (ΔABC内角和为180°)

=180°-90°+1/2∠A

=90°+1/2∠A.

①当已知∠ABC与∠ACB时,直接由角平分线与ΔPBC内角和可求.

②∠A=50°,则∠P=90°+1/2×50°=115°,

当∠A=110°,∠P=90°+1/2×110°=145°,