Question

已知：如图，∠BAP+∠APD =，∠1 =∠2.求证：∠E =∠F．

　

                                                                        

Answer 1

证明：∵ ∠BAP+∠APD = 180°，

　　∴ AB∥CD.

　　∴ ∠BAP =∠APC.

　　又∵ ∠1 =∠2,

　　∴ ∠BAP−∠1 =∠APC−∠2.

　　即∠EAP =∠APF.

　　∴ AEF∥P.

　　∴ ∠E =∠F.