Đặt BH = x (x > 0) => BC = (x + 6,4)
Có: AB2 = BH.BC => 62 = x.(x + 6,4) => 36 = x2 + 6,4x => x2 + 6,4x - 36 = 0
=> (x + 10)(5x - 18) = 0
=> x + 10 = 0 => x = -10 (loại)
hoặc 5x - 18 = 0 => x = 18/5 (nhận)
=> BH = 18/5cm , BC = 18/5 + 6,4 = 10cm
Ta có: AC2 = HC.BC = 6,4 . 10 = 64 => AC = 8cm
\(\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}=\frac{1}{6^2}+\frac{1}{8^2}=\frac{25}{576}\Rightarrow AH=\sqrt{\frac{576}{25}}=\frac{24}{5}cm\)
Vậy BC = 10cm , BH = 18/5cm , AH = 24/5cm , AC = 8cm