$AH=4BH$
Áp dụng định lý Pytago trong tam giác $AHB$
$AB^2=AH^2+HB^2=(4HB)^2+HB^2=17HB^2$
$⇒HB^2=\dfrac{AB^2}{17}=\dfrac{(2\sqrt{5})^2}{17}=\dfrac{20}{17}$
$⇒HB=\dfrac{2\sqrt{85}}{17}cm$
$AH=4HB=4.\dfrac{2\sqrt{85}}{17}=\dfrac{8\sqrt{85}}{17}cm$
$AH^2=HB.HC$
$⇒HC=\dfrac{AH^2}{HB}=\dfrac{\bigg(\dfrac{8\sqrt{85}}{17}\bigg)^2}{\dfrac{2\sqrt{85}}{17}}=\dfrac{32\sqrt{85}}{17}cm$
$BC=HB+HC=\dfrac{2\sqrt{85}}{17}+\dfrac{32\sqrt{85}}{17}=2\sqrt{85}$
$S_{∆ABC}=\dfrac{1}{2}AH.BC=\dfrac{1}{2}.\dfrac{8\sqrt{85}}{17}.2\sqrt{85}=40cm^2$
