a: \(HC=\dfrac{AH^2}{HB}=\dfrac{16}{3}\left(cm\right)\)
BC=BH+CH=16/3+3=25/3(cm)
\(AB=\sqrt{3\cdot\dfrac{25}{3}}=5\left(cm\right)\)
\(AC=\sqrt{\dfrac{16}{3}\cdot\dfrac{25}{3}}=\dfrac{20}{3}\left(cm\right)\)
b: Xét ΔADI có HB//ID
nên AH/HI=AB/BD
=>AH=HI
mà AH=1/2HE
nên HE=2HI
=>HI=IE
\(\tan IED=\dfrac{ID}{IE}=\dfrac{2\cdot HB}{AH}=\dfrac{2\cdot3}{4}=\dfrac{3}{2}\)