Question

问一道数学题，写解题过程

已知等腰三角形abc中，∠bac=90°，d是ac中点ae垂直bd交于e,交bc于f,连接df,求证：∠adb=∠cdf

Answer 1

证明：  如图：过点C作AC的垂线交AF的延长线于K

             由题得：∠1=∠3 （同角的余角相等）

               BA=AC

                      ∠BAD=∠ACK=90°

                 ∴ △BAD≌△ACK （角、边、角）

            AD=CK    ∠4=∠6

                    ∵ AD=CD

                    ∴CK=CD

                 又∵ ∠DCF=∠KCF=45°

                          CF=CF

                   ∴△DCF ≌△KCF （边、角、边）（“≌ 表示“全等于”）

                ∴ ∠5=∠6

                          ∴∠4=∠5

                   即：∠ADB=∠CDF

                                     《证毕》！