Giải thích các bước giải:

b.Xét $\Delta ABM,\Delta AMC$ có:

Chung $\hat A$

$\widehat{AMB}=\widehat{MCB}=\widehat{MCA}$

$\to \Delta ABM\sim\Delta AMC(g.g)$

$\to \dfrac{AB}{AM}=\dfrac{AM}{AC}$

$\to AM^2=AB.AC$

Mà $\Delta AMO$ vuông tại $M, MH\perp AO$

$\to AH.AO=AM^2$

$\to AH.AO=AB.AC$

$\to \dfrac{AH}{AB}=\dfrac{AC}{AO}$

$\to \Delta AHB\sim\Delta ACO(c.g.c)$

$\to \widehat{AHB}=\widehat{ACO}$

c.Ta có:

$\sin\widehat{MAO}=\dfrac{OM}{OA}=\dfrac12$

$\to \widehat{AMO}=30^o$

$\to \widehat{MAN}=2\widehat{MAO}=60^o$

Do $AM=AN$

$\to \Delta AMN$ đều

Ta có: $AM=\sqrt{AO^2-OM^2}=R\sqrt3$

$\to S_{AMN}=\dfrac{(R\sqrt3)^2\sqrt3}4=\dfrac{3\sqrt{3}R^2}{4}$