Question

如图,BP是∠ABC的平分线,DP是∠CDA的平分线,BO与DP交于P,若∠A=44°,∠C=56°,求∠P的大小

Answer 1

设 PB与AD相交于点E,PD与BC相交于点F.∠PBA=∠PBC=α,∠PDA=∠PDC=β

所以 ∠PEA=∠A+α=∠P+β,

∠PFC=∠C+β=∠P+α

两式相加,得 ∠A+∠C+α+β = 2∠P+α+β

所以 2∠P = ∠A+∠C = 44°+56° = 100°

故 ∠P = 50°