Bài 2:
a) $∆ABH \sim ∆CBA \, (\widehat{C}: \,chung)$
$\Rightarrow AB^2 = BH.BC$
$\Rightarrow AB^2 = BH.(BH+CH)$
mà $\dfrac{BH}{CH} = \dfrac{9}{16}$
$\Rightarrow BH = \dfrac{9}{16}CH$
$\Rightarrow BC = BH + CH = \dfrac{9}{16}CH +CH = \dfrac{25}{16}CH$
Do đó $AB^2 = \dfrac{9}{16}CH.\dfrac{25}{16}CH = \dfrac{25.9}{16^2}CH^2$
$\Rightarrow CH = \dfrac{16}{5} \,cm$
$\Rightarrow BC = \dfrac{25}{16}CH = 5 \, cm$
$\Rightarrow AC = \sqrt{BC^2 - AB^2} = 4 \, cm$
b) Ta có: $\dfrac{AB}{AC} = \dfrac{1}{2}$
$\Rightarrow AC = 2AB$
Bên cạnh đó: $AB.AC = AH.BC = 2.S_{ABC}$
$\Rightarrow AH = \dfrac{AB.AC}{BC}$
$\Rightarrow AH^2 = \dfrac{(AB.AC)^2}{AB^2 + AC^2} = \dfrac{4AB^2}{5}$
$\Rightarrow AB = \dfrac{5AH^2}{4} = 2\sqrt{5} \, cm$
$\Rightarrow AC = 2AB = 4\sqrt{5} \, cm$
$\Rightarrow BC =\dfrac{AB.AC}{AH} =\dfrac{2\sqrt{5}.\sqrt{5}}{4} = 10 \, cm$
