Câu hỏi của Lê Vũ Anh Thư

Question

Cho hình vuông ABCD. Trên tia đối tia của CD lấy điểm N. Đường thẳng AN cắt cạnh BC tại M. CMR:

a. AB² = BM . DN

b. $\dfrac{1}{AD^2}=\dfrac{1}{AM^2}+\dfrac{1}{AN^2}$...

Pham Van Hung · Accepted Answer

a,$\Delta ABM\infty\Delta NDA\left(g.g\right)\Rightarrow\frac{AB}{ND}=\frac{BM}{DA}\Rightarrow AB^2=BM.DN$ (vì AB = AD)

b, Ta có: $\frac{NM}{NA}=\frac{MC}{AD}\Rightarrow\frac{AD}{AN}=\frac{MC}{MN}$

$\frac{CN}{AB}=\frac{MN}{AM}\Rightarrow\frac{CN}{AD}=\frac{MN}{AM}\Rightarrow\frac{AD}{AM}=\frac{CN}{MN}$

Vậy $\left(\frac{AD}{AM}\right)^2+\left(\frac{AD}{AN}\right)^2=\left(\frac{CN}{MN}\right)^2+\left(\frac{MC}{MN}\right)^2=\frac{MC^2+CN^2}{MN^2}=1$

$\Rightarrow\frac{1}{AD^2}=\frac{1}{AM^2}+\frac{1}{AN^2}$

Câu trả lời:

a,$\Delta ABM\infty\Delta NDA\left(g.g\right)\Rightarrow\frac{AB}{ND}=\frac{BM}{DA}\Rightarrow AB^2=BM.DN$ (vì AB = AD)

b, Ta có: $\frac{NM}{NA}=\frac{MC}{AD}\Rightarrow\frac{AD}{AN}=\frac{MC}{MN}$

$\frac{CN}{AB}=\frac{MN}{AM}\Rightarrow\frac{CN}{AD}=\frac{MN}{AM}\Rightarrow\frac{AD}{AM}=\frac{CN}{MN}$

Vậy $\left(\frac{AD}{AM}\right)^2+\left(\frac{AD}{AN}\right)^2=\left(\frac{CN}{MN}\right)^2+\left(\frac{MC}{MN}\right)^2=\frac{MC^2+CN^2}{MN^2}=1$

$\Rightarrow\frac{1}{AD^2}=\frac{1}{AM^2}+\frac{1}{AN^2}$