Question

已知四邊形ABCD，AD∥BC，連接BD.

(1)小明說：“若添加條件BD²＝BC²＋CD²，則四邊形ABCD是矩形”.你認為小明的說法是否正確，若正確請說明理由，若不正確，請舉出一個反例說明；

(2)若BD平分∠ABC，∠DBC＝∠BDC，tan∠DBC＝1.求證：四邊形ABCD是正方形.

Answer 1

 (1)解：不正確.                                                        (1分)

如圖作(直角)梯形ABCD，                                              (2分)

　 使得AD∥BC，∠C＝90°.

連接BD，則有BD²＝BC²＋CD².                                         (3分)

而四邊形ABCD是直角梯形而不是矩形.                                   (4分)

 (2)證明：如圖，

　∵tan∠DBC＝1，

∴∠DBC＝45°.                                                        (5分)

∵∠DBC＝∠BDC，

∴∠BDC＝45°.

則BC＝DC.                                                           (6分)

法1：∵BD平分∠ABC，

∴∠ABD＝45°，∴∠ABD＝∠BDC.

∴AB∥DC.

∴四邊形ABCD是平行四邊形.                                           (7分)

又∵∠ABC＝45°＋45°＝90°，

∴四邊形ABCD是矩形.                                                (8分)

∵BC＝DC，

∴四邊形ABCD是正方形.                                              (9分)

法2：∵BD平分∠ABC，∠BDC＝45°，∴∠ABC＝90°.

∵∠DBC＝∠BDC＝45°，∴∠BCD＝90°.

∵AD∥BC，

∴∠ADC＝90°.                                                        (7分)

∴四邊形ABCD是矩形.                                                 (8分)

又∵BC＝DC

∴四邊形ABCD是正方形.                                               (9分)

法3：∵BD平分∠ABC，∴ ∠ABD＝45°.

∴∠BDC＝∠ABD.

∵AD∥BC，∴∠ADB＝∠DBC.

∵BD＝BD，

∴△ADB≌△CBD.

∴AD＝BC＝DC＝AB.                                                  (7分)

∴四邊形ABCD是菱形.                                                (8分)

又∵∠ABC＝45°＋45°＝90°，

∴ 四邊形ABCD是正方形.                                              (9分)