a) Đặt $AB = AC = x \, (x > 0)$
Ta có: $BC = 2CH = 2\sqrt{AC^2 - AH^2}$
$BC.AH = BK.AC = 2S_{ABC}$
$\Leftrightarrow 2\sqrt{x^2 - 16^2}.16 = 19,2x$
$\Leftrightarrow 4.16^2(x^2 - 16^2) = 19,2^2x^2$
$\Leftrightarrow x^2 = 400$
$\Rightarrow x = 20 \, cm$
Vậy $AB = AC = 20 \, cm$
b) Gọi $M$ là trung điểm $AC$
$\Rightarrow OM \perp AC;\, MA = MC = \dfrac{1}{2}AC = 10 \, cm$
Ta có: $∆AOM\sim ∆ACH \, (g.g)$
$\Rightarrow \dfrac{AO}{AC} = \dfrac{AM}{AH}$
$\Rightarrow AO = \dfrac{AC.AM}{AH} = \dfrac{20.10}{16} = \dfrac{25}{2} \, cm$
$\Rightarrow OH = AH - AO = 16 - \dfrac{25}{2} = \dfrac{7}{2} \, cm$
