Question

[HoangDack VN]

b1: cho Δabc lay MϵBC, ME//AC, MD//AB(DϵAC). Tinh AE/AB + AD/AC = ?
b2: cho Δabc, lay DϵAC sao cho AD/CD=1/2. Goi M la trung diem cua BD. Tia AM cat BC tai E. DF//AE(FϵBC). a, tinh CE/EF
b, tinh CE
Hua danh gia 5* lun

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 1: 

Xét ΔABC có

MD//AB

nên \(\dfrac{AD}{AC}=\dfrac{BM}{BC}\)

Xét ΔABC có 

ME//AC

nên \(\dfrac{AE}{AB}=\dfrac{CM}{CB}\)

Ta có: \(\dfrac{AE}{AB}+\dfrac{AD}{AC}\)

\(=\dfrac{BM}{BC}+\dfrac{CM}{BC}\)

=1

Answer 2

[Shauna]