Giải thích các bước giải:
a.Ta có $\widehat{AHB}=\widehat{AHC},\widehat{BAH}=90^o-\widehat{HAC}=\widehat{ACH}$
$\to\Delta ABH\sim\Delta CAH(g.g)$
Mà $M,N$ là trung điểm $BH, AH$
$\to \Delta BMA\sim\Delta ANC$
b.Từ câu a $\to\widehat{ACN}=\widehat{BAM}$
$\to\widehat{ACN}+\widehat{CAM}=\widehat{BAM}+\widehat{CAM}$
$\to \widehat{CAO}+\widehat{OAC}=90^o$
$\to AO\perp CO\to AM\perp CN$
c.Vì $AD$ là phân giác góc $A\to\dfrac{DB}{DC}=\dfrac{AB}{AC}$
Từ câu a $\to \dfrac{AB}{AC}=\dfrac{AH}{CH}$
$\to\dfrac{DB}{DC}=\dfrac{AH}{CH}$
$\to AH.DC=BD.HC$
d.Ta có $BC=10\to BC^2=100\to AB^2+AC^2=100$
Trên tia đối của tia $DA$ lấy $E$ sao cho $\widehat{DCE}=\widehat{DAC}=\widehat{BAD}$
Mà $\widehat{ADB}=\widehat{CDE}$
$\to\Delta ADB\sim\Delta CDE(g.g)$
$\to\dfrac{DA}{DC}=\dfrac{DB}{DE}$
$\to DA.DE=DB.DC$
Mặt khác $\widehat{ABD}=\widehat{DEC}\to \widehat{ABD}=\widehat{AEC}$
Mà $\widehat{BAD}=\widehat{BAC}=\widehat{EAC}$
$\to\Delta ABD\sim\Delta AEC(g.g)$
$\to\dfrac{AB}{AE}=\dfrac{AD}{AC}$
$\to AB.AC=AD.AE$
$\to AB.AC-DB.DC=AD.AE-AD.DE=AD^2$
$\to AB.AC-DB.(BC-DB)=AD^2$
$\to AB.AC-\dfrac23AB.(10-\dfrac23AB)=(\dfrac23 AB)^2$
$\to AB.AC-\dfrac{20}{3}AB+\dfrac49 AB^2=\dfrac49AB^2$
$\to AB.AC=\dfrac{20}{3}AB$
$\to AC=\dfrac{20}{3}$
$\to AB=\sqrt{BC^2-AC^2}=\sqrt{10^2-(\dfrac{20}{3})^2}=\dfrac{10\sqrt{5}}{3}$
$\to AB+AC=\dfrac{10\sqrt{5}}{3}+\dfrac{20}{3}$
$\to S_{ABC}=\dfrac12AB\cdot AC =\dfrac12\cdot\dfrac{10\sqrt{5}}{3}\cdot \dfrac{20}{3}=\dfrac{100\sqrt{5}}{9}$
$\to \dfrac12AH\cdot BC=\dfrac{100\sqrt{5}}{9}$
$\to \dfrac12 AH\cdot 10=\dfrac{100\sqrt{5}}{9}$
$\to AH=\dfrac{20\sqrt{5}}{9}$
