$\color{cadetblue}{\large{@Sandy}}$

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$\color{darkgoldenrod}{a,}\ \text{Xét}\ \triangle ABC\ \text{và}\ \triangle HBA\ \text{có:}$

$\hspace{0.65cm}\ \widehat{B}\ \text{chung}$

$\hspace{0.65cm}\ \widehat{BAC}=\widehat{BHA}\ (=90^\circ)$

$\Rightarrow\ \triangle ABC\ \backsim\ \triangle HBA\ \text{(g - g)}$

$\Rightarrow\ \dfrac{AB}{BH}=\dfrac{BC}{AB}\ \text{(2 cạnh t/ứ)}$

$\Leftrightarrow\ AB^2=BH.BC\ \text{(đpcm)}$

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$\color{darkgoldenrod}{b,}\ AB^2=BH.BC\ \text{(cmt)}$

$\hspace{0.65cm}\ \text{Mà}\ AB=3cm\ \text{(gt)}$

$\hspace{1.37cm}\ BC=5cm\ \text{(gt)}$

$\Rightarrow\ 3^2=5BH$

$\Leftrightarrow\ BH=1,8\ (cm)$

$\hspace{0.65cm}\ \text{Xét}\ \triangle HBA\ \text{vuông tại}\ H\ \text{có:}$

$\hspace{0.65cm}\ BH^2+AH^2=AB^2\ \text{(định lí Pythagoras)}$

$\hspace{0.65cm}\ \text{Mà}\ AB=3cm\ \text{(gt)}$

$\hspace{1.37cm}\ BH=1,8cm\ \text{(cmt)}$

$\Rightarrow\ (1,8)^2+AH^2=3^2$

$\Leftrightarrow\ 3,24+AH^2=9$

$\Leftrightarrow\ AH^2=5,76$

$\Leftrightarrow\ AH=2,4\ (TMĐK)$

$\hspace{0.65cm}\ \text{Vậy}\ BH=1,8cm$

$\hspace{1.5cm}\ AH=2,4cm$

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$\color{darkgoldenrod}{c,}\ \text{Xét}\ \triangle AHC\ \text{có:}$

$\hspace{0.35cm}\ \left.\begin{matrix} I\ \text{là trung điểm}\ HC\ \text{(gt)}\\K\ \text{là trung điểm}\ AH\ \text{(gt)} \end{matrix}\right\}\ \Rightarrow\ IK\ \text{là đường TB}\ \triangle AHC$

$\Rightarrow\ IK//AC\ \text{(t/c ĐTB)}$

$\hspace{0.65cm}\ AC\bot AB$

$\Rightarrow\ IK\bot AB$

$\hspace{0.65cm}\ \text{Xét}\ \triangle AIB\ \text{có:}$

$\hspace{0.65cm}\ IK\bot AB\ \text{(cmt)}$

$\hspace{0.65cm}\ AH\bot BC\ \text{(gt)}$

$\hspace{0.65cm}\ IK\cap AH\ \text{tại}\ K$

$\Rightarrow\ K\ \text{là trực tâm}\ \triangle AIB$

$\Rightarrow\ BK\bot AI\ \text{(đpcm)}$