a: AB=AC/2=2cm
\(BC=\sqrt{2^2+4^2}=2\sqrt{5}\)
\(AH=\dfrac{AB\cdot AC}{BC}=\dfrac{2\cdot4}{2\sqrt{5}}=\dfrac{4\sqrt{5}}{5}\left(cm\right)\)
\(BH=\dfrac{AB^2}{BC}=\dfrac{4}{4\sqrt{5}}=\dfrac{\sqrt{5}}{5}\left(cm\right)\)
\(CH=\dfrac{AC^2}{CB}=\dfrac{4^2}{4\sqrt{5}}=\dfrac{4\sqrt{5}}{\sqrt{5}}\left(cm\right)\)
\(\tan B=\dfrac{AC}{AB}=\dfrac{4}{2}=2\)
\(\tan C=\dfrac{AB}{AC}=\dfrac{1}{2}\)
b: \(AB=\dfrac{2}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3}\left(cm\right)\)
\(BC=\sqrt{2^2+\left(\dfrac{2}{\sqrt{3}}\right)^2}=\dfrac{4\sqrt{3}}{3}\left(cm\right)\)
\(BH=\dfrac{AB^2}{BC}=\dfrac{\sqrt{3}}{3}\left(cm\right)\)
\(CH=\dfrac{AC^2}{BC}=\sqrt{3}\left(cm\right)\)
\(AH=\sqrt{BH\cdot CH}=\sqrt{\dfrac{3}{3}}=1\left(cm\right)\)
c: \(AC=2\sqrt{3}\left(cm\right)\)
\(BC=\sqrt{12+12}=2\sqrt{6}\left(cm\right)\)
\(AH=\dfrac{AB\cdot AC}{BC}=\dfrac{12}{2\sqrt{6}}=\sqrt{6}\left(cm\right)\)
\(BH=\dfrac{AB^2}{BC}=\dfrac{12}{2\sqrt{6}}=\sqrt{6}\left(cm\right)\)
\(CH=\sqrt{6}\left(cm\right)\)
