Question

已知：等边△ABC

（1）如图1，P为等边△ABC外一点，且∠BPC=120°．试猜想线段BP、PC、AP之间的数量关系，并证明你的猜想；
（2）如图2，P为等边△ABC内一点，且∠APD=120°．求证：PA+PD+PC＞BD．

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                 

Answer 1

猜想：AP=BP+PC，
（1）证明：延长BP至E，使PE=PC，连接CE，
∵∠BPC=120°，
∴∠CPE=60°，又PE=PC，
∴△CPE为等边三角形，
∴CP=PE=CE，∠PCE=60°，
∵△ABC为等边三角形，
∴AC=BC，∠BCA=60°，
∴∠ACB=∠PCE，
∴∠ACB+∠BCP=∠PCE+∠BCP，
即：∠ACP=∠BCE，
∴△ACP≌△BCE，
∴AP=BE，
∵BE=BP+PE，
∴AP=BP+PC．   

（2）证明：
在AD外侧作等边△AB′D，
则点P在三角形ADB′外，
∵∠APD=120°∴由（1）得PB′=AP+PD，
在△PB′C中，有PB′+PC＞CB′，
∴PA+PD+PC＞CB′，
∵△AB′D、△ABC是等边三角形，
∴AC=AB，AB′=AD，
∠BAC=∠DAB′=60°，
∴∠BAC+∠CAD=∠DAB′+∠CAD，
即：∠BAD=∠CAB′，
∴△AB′C≌△ADB，
∴CB′=BD，
∴PA+PD+PC＞BD．