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Kết quả tính $BN$

Cho hình chóp $S.ABC$ với $SA = AB = BC = a$:

 $AC = a\sqrt{2}$

$SC = \sqrt{SA^2 + AC^2} = a\sqrt{3}$

$SB = \sqrt{SA^2 + AB^2} = a\sqrt{2}$

$\triangle MNC \sim \triangle SAC \Rightarrow NC = \frac{MC \cdot AC}{SC} = \frac{\frac{a\sqrt{2}}{2} \cdot a\sqrt{2}}{a\sqrt{3}} = \frac{a}{\sqrt{3}}$

$\cos \widehat{SCB} = \frac{BC^2 + SC^2 - SB^2}{2 \cdot BC \cdot SC} = \frac{a^2 + 3a^2 - 2a^2}{2 \cdot a \cdot a\sqrt{3}} = \frac{1}{\sqrt{3}}$

$BN^2 = BC^2 + NC^2 - 2BC \cdot NC \cdot \cos \widehat{SCB}$

$BN^2 = a^2 + \frac{a^2}{3} - 2a \cdot \frac{a}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}} = a^2 + \frac{a^2}{3} - \frac{2a^2}{3} = \frac{2a^2}{3}$

Đáp số: $BN = \frac{a\sqrt{6}}{3}$