Giải thích các bước giải:
Ta có:
$S_{ABC}=\dfrac12AB\cdot AC$
$\to \dfrac12AB\cdot AC=20$
$\to \dfrac12\cdot 5\cdot AC=20$
$\to AC=8$
$\to BC^2=AB^2+AC^2=5^2+8^2=89$
$\to BC=\sqrt{89}$
Mà $\dfrac12\cdot AH\cdot BC=S_{ABC}$
$\to \dfrac12\cdot AH\cdot \sqrt{89}=20$
$\to AH=\dfrac{40}{\sqrt{89}}$
$\to HB=\sqrt{AB^2-AH^2}=\sqrt{5^2-(\dfrac{40}{\sqrt{89}})^2}=\dfrac{25}{\sqrt{89}}$
$\to HC=BC-HB=\dfrac{84}{\sqrt{89}}$