Question

如圖，已知：⊙O的弦AB的延長線和切線EP交于點P，E為切點，連結EA、EB，過點P的一條直線交EA、EB于C、D，若EC=EＤ．

求證：(1)∠APC=∠CPE;

(2)PA·CE=AC·PE；

(3)ED²=AC·BＤ．

 

Answer 1

答案：
解析：

（1）∵EP為切線，∴∠PEB=∠A．∵EC=ED,∴∠ECD=∠EDC． ∵∠ECD=∠A+∠APC,∠EDC=∠PEB+∠CPE,∴∠APC=∠CPE． (2)∵∠APC=∠CPE, ∴由角平分線性質，得． ∴PA·CE=AC·PE． （3）∵∠APC=∠CPE． ∴                                                                                                                ①                                                                                                                    ② ∵PE為切線， ∴                                                                                                                       ③ 由①、②、③，得． 又EC=ED， ∴ED2=AC·BD．