cho tam giác ABC.Từ D trên cạnh AB, kẻ đoạn thẳng song song vs BC cắt AC tại E . trên tia đối của tia CA, lấy F sao cho CF=DB. gọi M là giao điểm của DF và BC CMR:DM.AB=AC.MF
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51 Tham khảo câu trả lời từ Trợ lý ảo Lazi Để chứng minh CMR: DM.AB = AC.MF, ta sẽ sử dụng định lí Menelaus trong tam giác ABC. Áp dụng định lí Menelaus trong tam giác ABC với đường chéo DF, ta có: (DM/MB) * (BC/CA) * (AE/ED) = 1 Vì DF || BC, nên theo định lí Thales, ta có: DM/MB = CF/FB Vì CF = DB (theo đề bài), nên ta có: DM/MB = DB/FB Thay vào biểu thức trên, ta có: (DB/FB) * (BC/CA) * (AE/ED) = 1 Đặt x = DB/FB, y = AE/ED, ta có: x * (BC/CA) * y = 1 Từ đó, ta suy ra: x = CA/BC * 1/y Vì CF = DB, nên ta có: CF = x * FB = (CA/BC) * (FB/y) Gọi G là giao điểm của CF và AB. Áp dụng định lí Menelaus trong tam giác ABC với đường chéo CF, ta có: (CG/GB) * (AB/AC) * (CF/FB) = 1 Thay vào biểu thức trên, ta có: (CG/GB) * (AB/AC) * ((CA/BC) * (FB/y))/(FB) = 1 Simplifying, we get: (CG/GB) * (AB/AC) * (CA/yBC) = 1 Từ đó, ta suy ra: (CG/GB) * (AB/AC) * (CA/BC) * (BC/yBC) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1 (CG/GB) * (AB/AC) * 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