-△ABC∼△HBA (g-g) \(\Rightarrow\dfrac{P_{ABC}}{P_{HBA}}=\dfrac{BC}{BA}=\dfrac{20}{12}=\dfrac{5}{3}\Rightarrow\dfrac{AB}{BC}=\dfrac{3}{5}\)
\(\Rightarrow AB=\dfrac{3}{5}BC\)
-△ABC vuông tại A có: \(AB^2+AC^2=BC^2\Rightarrow\dfrac{9}{25}BC^2+AC^2=BC^2\Rightarrow AC^2=\dfrac{16}{25}BC^2\Rightarrow AC=\dfrac{4}{5}BC\)
-△ABC∼△HAC (g-g) \(\Rightarrow\dfrac{P_{ABC}}{P_{HAC}}=\dfrac{BC}{AC}=\dfrac{BC}{\dfrac{4}{5}BC}=\dfrac{5}{4}\Rightarrow\dfrac{20}{P_{HAC}}=\dfrac{5}{4}\Rightarrow P_{HAC}=\dfrac{20.4}{5}=16\left(cm\right)\)