áp dụng HTL:
\(AH^2=BH\cdot HC\Leftrightarrow AH=\sqrt{BH\cdot HC}=\sqrt{5\cdot9}=3\sqrt{5}cm\)
\(AB=\sqrt{AH^2+BH^2}=\sqrt{\left(3\sqrt{5}\right)^2+5^2}=\sqrt{70}cm\left(Pytago\right)\)
\(AC=\sqrt{AH^2+HC^2}=\sqrt{\left(3\sqrt{5}\right)^2+9^2}=3\sqrt{14}cm\left(Pytago\right)\)
\(BC=\sqrt{AB^2+AC^2}=\sqrt{\left(\sqrt{70}\right)^2+\left(3\sqrt{14}\right)^2}=14cm\left(Pytago\right)\)
\(BC=BH+HC=14\left(cm\right)\)
Áp dụng HTL: \(\left\{{}\begin{matrix}AH=\sqrt{BH.HC}=3\sqrt{5}\left(cm\right)\\AB=\sqrt{BC.BH}=\sqrt{70}\left(cm\right)\\AC=\sqrt{BC.CH}=3\sqrt{14}\left(cm\right)\end{matrix}\right.\)
