∴∠PBC=
1 |
2 |
1 |
2 |
∴∠PBC+∠PCB=
1 |
2 |
1 |
2 |
∴∠P=180°-(∠PCB+∠PBC)=90°+
1 |
2 |
(2)∠ACE=∠A+∠ABC,
∵CP平分∠ACE,BP平分∠ABC,
∴∠PBC=
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
∴∠P=180°-(∠PBC+∠PCA+∠ACB)=
1 |
2 |
(3)∵∠DBC=∠A+∠ACB,∠ECB=∠A+∠ABC,
∴∠DBC+∠ECB=∠A+∠ACB+∠A+∠ABC,
∵BP、CP分别平分∠DBC和∠ECD,
∴∠PBC=
1 |
2 |
1 |
2 |
∴∠PBC+∠PCB=
1 |
2 |
∴∠P=180°-(∠PBC+∠PCB)=180°-
1 |
2 |
1 |
2 |
故答案为:(1)∠P=90°+
1 |
2 |
1 |
2 |