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Cho tam giác ABC

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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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

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(CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1

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(CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1

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(CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1

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(CG/GB) * (AB/AC) * (CA/BC) * (1/y) = 1

(CG/GB) * (AB/AC) * (

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