如图,已知∠DBA和∠ACE是△ABC的外角,求∠DBA+∠ACE与∠A的关系

∵∠ABD是ΔABC的外角,∴∠ABD=∠A+∠ACB,
∵∠ACE是ΔABC的外角,∴∠ACE=∠A+∠ABC,
∴∠ABD+∠ACE=∠A+(∠A+∠ABC+∠ACB)=180°+∠A.△ABC中,BP、CP分别平分∠ABC、∠ACB,∠A=80°,求∠BPC的度数∵BP、CP分别平分∠ABC、∠ACB，∴∠PBC=1/2ABC，∠PCB=1/2∠ACB，∴∠PBC+∠PCB=1/2(∠ABC+∠ACB)=1/2(180°-∠A)=90°1/2∠A，∴∠BPC=180°-(∠PBC+∠PCB)=180°-(90°-1/2∠A)=90°+1/2∠A，=130。=90°1/2∠A？∵BP、CP分别平分∠ABC、∠ACB，∴∠PBC=1/2ABC，∠PCB=1/2∠ACB，∴∠PBC+∠PCB=1/2(∠ABC+∠ACB)=1/2(180°-∠A)=90°-1/2∠A，△ABC中,BP,CP分别平分∠ABC,∠ACB,∠BPC=132°，求∠A的度数？