Đặt cạnh hình vuông là a
Dễ tính được: \(AN=\sqrt{AD^2+DN^2}=\dfrac{a\sqrt{5}}{2}\), \(MN=\sqrt{MC^2+CN^2}=\sqrt{\left(\dfrac{a}{2}\right)^2+\left(\dfrac{a}{2}\right)^2}=\dfrac{a\sqrt{2}}{2}\)
\(AM=\sqrt{AB^2+BM^2}=\sqrt{a^2+\left(\dfrac{a}{2}\right)^2}=\dfrac{a\sqrt{5}}{2}\)
Kẻ \(MK\perp AN\)
Ta chứng minh: \(cos\widehat{ANM}=\dfrac{AN^2+MN^2-AM^2}{2AN.NM}\) (1)
(1) \(\Leftrightarrow2AN.MN.cos\widehat{N}=AN^2+MN^2-AM^2\)
\(\Leftrightarrow2.AN.MN.\dfrac{KN}{MN}=\left(AK+KN\right)^2+MK^2+NK^2-MK^2-AK^2\)
\(\Leftrightarrow2.AN.KN=AK^2+2.AK.KN+KN^2+NK^2-AK^2\)
\(\Leftrightarrow2KN.AK-2AN.NK+2KN^2=0\)
\(\Leftrightarrow2KN\left(AK-AN+KN\right)=0\) \(\Leftrightarrow2.KN.0=0\) (lđ)
Từ (1) \(\Rightarrow cos\widehat{ANM}=\dfrac{\left(\dfrac{a\sqrt{5}}{2}\right)^2+\left(\dfrac{a\sqrt{2}}{2}\right)^2-\left(\dfrac{a\sqrt{5}}{2}\right)^2}{2.\left(\dfrac{a\sqrt{5}}{2}\right)\left(\dfrac{a\sqrt{2}}{2}\right)}\)\(=\dfrac{\sqrt{10}}{10}\)
Ý B
