a)

+ Ta có $\Delta ABC$ vuông tại $A$

Nên $B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}$ (Định lý Pitago)

$\Rightarrow AC=\sqrt{B{{C}^{2}}-A{{B}^{2}}}=\sqrt{{{7,5}^{2}}-{{4,5}^{2}}}=6cm$

+ Ta có ${{S}_{\Delta ABC}}=\dfrac{1}{2}AH.BC=\dfrac{1}{2}AB.AC$

Nên $AH.BC=AB.AC\Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{4,5.6}{7,5}=3,6cm$

+ Ta có $AD$ là phân giác $\widehat{BAC}$

Nên $\dfrac{AB}{BD}=\dfrac{AC}{CD}=\dfrac{AB+AC}{BD+CD}=\dfrac{AB+AC}{BC}=\dfrac{4,5+6}{7,5}=\dfrac{7}{5}$

Với $\dfrac{AB}{BD}=\dfrac{7}{5}\Rightarrow BD=\dfrac{5}{7}AB=\dfrac{5}{7}\cdot 4,5=\dfrac{45}{14}cm$

Với $\dfrac{AC}{CD}=\dfrac{7}{5}\Rightarrow CD=\dfrac{5}{7}AC=\dfrac{5}{7}\cdot 6=\dfrac{30}{7}cm$

b)

+ Ta có $\Delta AHB$ vuông tại $H$

Nên $A{{B}^{2}}=B{{H}^{2}}+A{{H}^{2}}$ (Định lý Pitago)

$\Rightarrow BH=\sqrt{A{{B}^{2}}-A{{H}^{2}}}=\sqrt{{{4,5}^{2}}-{{3,6}^{2}}}=2,7cm$

+ ${{S}_{\Delta AHD}}=\dfrac{1}{2}AH.HD=\dfrac{1}{2}AH\left( BD-BH \right)=\dfrac{1}{2}\cdot 3,6\left( \dfrac{45}{14}-2,7 \right)=\dfrac{162}{175}c{{m}^{2}}$

c)

+ Có $AH.BC=AB.AC$

$\Rightarrow A{{H}^{2}}.B{{C}^{2}}=A{{B}^{2}}.A{{C}^{2}}$

$\Rightarrow \dfrac{1}{A{{H}^{2}}}=\dfrac{B{{C}^{2}}}{A{{B}^{2}}.A{{C}^{2}}}$

$\Rightarrow \dfrac{1}{A{{H}^{2}}}=\dfrac{A{{B}^{2}}+A{{C}^{2}}}{A{{B}^{2}}.A{{C}^{2}}}$

$\Rightarrow \dfrac{1}{A{{H}^{2}}}=\dfrac{1}{A{{B}^{2}}}+\dfrac{1}{A{{C}^{2}}}$

+ Áp dụng hệ quả định lý Ta-let:

Có $HM\bot AB,AC\bot AB$

Nên $HM//AC\Rightarrow \dfrac{MH}{AC}=\dfrac{BH}{BC}$

Có $HN\bot AB,AC\bot AB$

Nên $HN//AB\Rightarrow \dfrac{HN}{AB}=\dfrac{CH}{BC}$

$\Rightarrow \dfrac{HN}{AB}+\dfrac{MH}{AC}=\dfrac{BH+CH}{BC}=\dfrac{BC}{BC}=1$