Question

如图，已知：⊙O的弦AB的延长线和切线EP交于点P，E为切点，连结EA、EB，过点P的一条直线交EA、EB于C、D，若EC=EＤ．

求证：(1)∠APC=∠CPE;

(2)PA·CE=AC·PE；

(3)ED²=AC·BＤ．

 

Answer 1

答案：
解析：

（1）∵EP为切线，∴∠PEB=∠A．∵EC=ED,∴∠ECD=∠EDC． ∵∠ECD=∠A+∠APC,∠EDC=∠PEB+∠CPE,∴∠APC=∠CPE． (2)∵∠APC=∠CPE, ∴由角平分线性质，得． ∴PA·CE=AC·PE． （3）∵∠APC=∠CPE． ∴                                                                                                                ①                                                                                                                    ② ∵PE为切线， ∴                                                                                                                       ③ 由①、②、③，得． 又EC=ED， ∴ED2=AC·BD．