Ta có: BC2 = AB2 + AC2 \(\Rightarrow BC=\sqrt{AB^2+AC^2}=\sqrt{12^2+16^2}=20cm\)
\(AB^2=HB.BC\Rightarrow HB=\frac{AB^2}{BC}=\frac{12^2}{20}=\frac{36}{5}=7,2cm\)
\(AC^2=HC.BC\Rightarrow HC=\frac{AC^2}{BC}=\frac{16^2}{20}=\frac{64}{5}=12,8cm\)
Vì AD là phân giác góc BAC nên ta có :
\(\frac{DB}{DC}=\frac{AB}{AC}=\frac{12}{16}=\frac{3}{4}\Rightarrow DC=\frac{4}{7}BC=\frac{4}{7}.20=\frac{80}{7}cm\)
=> HD = BC - (HB + DC) \(=20-\left(7,2+\frac{80}{7}\right)=\frac{48}{35}cm\)
Vậy HB = 7,2cm ; HC = 12,8cm ; HD = 48/35cm