Giải thích các bước giải:
a.Ta có $\Delta ABC$ vuông tại $A,AH\perp BC$
$\to\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC}^2$
$\to\dfrac{1}{6^2}=\dfrac{1}{(4\sqrt{3})^2}+\dfrac{1}{AC}^2$
$\to AC=12$
$\to BC^2=AB^2+AC^2=(4\sqrt{3})^2+12^2=192$
$\to BC=8\sqrt{3}$
$\to \sin\widehat{ABC}=\dfrac{AC}{BC}=\dfrac{12}{8\sqrt{3}}=\dfrac{\sqrt3}{2}$
$\to\widehat{ABC}=60^o\to\widehat{ACB}=90^o-\widehat{ABC}=30^o$
b.Ta có $HM\perp AB\to\widehat{AMH}=\widehat{AHB}=90^o$
Mà $\widehat{MAH}=\widehat{BAH}$
$\to\Delta AMH\sim\Delta AHB(g.g)$
$\to\dfrac{AM}{AH}=\dfrac{AH}{AB}$
$\to AM.AB=AH^2$
Tương tự $AN.AC=AH^2$
$\to AM.AB=AN.AC$