$\hspace{0.65cm}\ \textbf{GT}\hspace{0.6cm}\triangle\ \text{nhọn ABC}$

$\hspace{2.2cm}\text{Đường cao AD, BE, CF}$

$\hspace{2.2cm}\text{Trực tâm H}$

$\\$

$\hspace{0.65cm}\ \textbf{KL}\hspace{0.66cm}\text{a)}\ \triangle\text{ABD}\backsim\triangle\text{AHF và AF.AB = AH.AD}$

$\hspace{2.2cm}\text{b) AF.AB = AE.AC và}\ \triangle\text{AEF}\backsim\triangle\text{ABC}$

$\hspace{2.2cm}\text{c) FC là phân giác}\ \mathrm{\widehat{EFD}}\ \text{và}\ \mathrm{BC^2=BH.BE+CH.CF}$

$\\$$\\$$\\$

$\color{darkgoldenrod}{a)}\ \ \ $Xét $\triangle ABD$ và $\triangle AHF$ có:

$\hspace{0.65cm}\ \widehat{BAD}$ chung

$\hspace{0.65cm}\ \widehat{ADB}=\widehat{AFH}\ (=90^\circ )$

$\Rightarrow\ \triangle ABD\ \backsim\ \triangle AHF\ (g-g)\ (đpcm)$

$\Rightarrow\ \dfrac{AB}{AH}=\dfrac{AD}{AF}$

$\Leftrightarrow\ AF.AB=AH.AD\ \ (đpcm)$

$\\$$\\$

$\color{darkgoldenrod}{b)}\ \ \ $Xét $\triangle ACD$ và $\triangle AHE$ có:

$\hspace{0.65cm}\ \widehat{CAD}$ chung

$\hspace{0.65cm}\ \widehat{ADC}=\widehat{AEH}\ (=90^\circ )$

$\Rightarrow\ \triangle ACD\ \backsim\ \triangle AHE\ (g-g)$

$\Rightarrow\ \dfrac{AC}{AH}=\dfrac{AD}{AE}$

$\Leftrightarrow\ AE.AC=AH.AD$

$\hspace{0.65cm}\ $Mà $AF.AB=AH.AD\ \ (cmt)$

$\Rightarrow\ AF.AB=AE.AC\ \ (đpcm)$

$\Leftrightarrow\ \dfrac{AF}{AE}=\dfrac{AC}{AB}$

$\hspace{0.65cm}\ $Xét $\triangle AEF$ và $\triangle ABC$ có:

$\hspace{0.65cm}\ \widehat{BAC}$ chung

$\hspace{0.65cm}\ \dfrac{AF}{AE}=\dfrac{AC}{AB}\ (cmt)$

$\Rightarrow\ \triangle AEF\ \backsim\ \triangle ABC\ (c-g-c)\ (đpcm)$

$\\$$\\$

$\color{darkgoldenrod}{c)}\ \ \ $Xét $\triangle CHD$ và $\triangle CBF$ có:

$\hspace{0.65cm}\ \widehat{BCF}$ chung

$\hspace{0.65cm}\ \widehat{CDH}=\widehat{CFB}\ (=90^\circ )$

$\Rightarrow\ \triangle CHD\ \backsim\ \triangle CBF\ (g-g)$

$\Rightarrow\ \dfrac{CH}{CB}=\dfrac{CD}{CF}$

$\Leftrightarrow\ CH.CF=CB.CD\ \ \ \ \ \ \ \color{red}{(5)}$

$\Leftrightarrow\ \dfrac{CH}{CD}=\dfrac{CB}{CF}$

$\hspace{0.65cm}\ $Xét $\triangle CDF$ và $\triangle CHB$ có:

$\hspace{0.65cm}\ \widehat{BCF}$ chung

$\hspace{0.65cm}\ \dfrac{CH}{CD}=\dfrac{CB}{CF}\ (cmt)$

$\Rightarrow\ \triangle CDF\ \backsim\ \triangle CHB\ (c-g-c)$

$\Rightarrow\ \widehat{CFD}=\widehat{CBH}\ \ \ \ \ \ \ \color{red}{(1)}$

$\hspace{0.65cm}\ $Xét $\triangle CHE$ và $\triangle CAF$ có:

$\hspace{0.65cm}\ \widehat{ACF}$ chung

$\hspace{0.65cm}\ \widehat{CEH}=\widehat{CFA}\ (=90^\circ )$

$\Rightarrow\ \triangle CHE\ \backsim\ \triangle CAF\ (g-g)$

$\Rightarrow\ \dfrac{CH}{CA}=\dfrac{CE}{CF}$

$\Leftrightarrow\ CH.CF=CA.CE$

$\Leftrightarrow\ \dfrac{CH}{CE}=\dfrac{CA}{CF}$

$\hspace{0.65cm}\ $Xét $\triangle CEF$ và $\triangle CHA$ có:

$\hspace{0.65cm}\ \widehat{ACF}$ chung

$\hspace{0.65cm}\ \dfrac{CH}{CE}=\dfrac{CA}{CF}\ (cmt)$

$\Rightarrow\ \triangle CEF\ \backsim\ \triangle CHA\ (c-g-c)$

$\Rightarrow\ \widehat{CFE}=\widehat{CAH}\ \ \ \ \ \ \ \color{red}{(2)}$

$\hspace{0.65cm}\ $Xét $\triangle CAD$ và $\triangle CBE$ có:

$\hspace{0.65cm}\ \widehat{ACB}$ chung

$\hspace{0.65cm}\ \widehat{ADC}=\widehat{BEC}\ (=90^\circ )$

$\Rightarrow\ \triangle CAD\ \backsim\ \triangle CBE\ (g-g)$

$\Rightarrow\ \widehat{CAH}=\widehat{CBH}\ \ \ \ \ \ \ \color{red}{(3)}$

$\hspace{0.65cm}\ $Từ $(1),(2),(3)\ \Rightarrow\ \widehat{CFD}=\widehat{CFE}$

$\Rightarrow\ FC$ là phân giác $\widehat{EFD}\ \ (đpcm)$

$\\$$\\$

$\hspace{0.65cm}\ $Xét $\triangle BHD$ và $\triangle BCE$ có:

$\hspace{0.65cm}\ \widehat{EBC}$ chung

$\hspace{0.65cm}\ \widehat{BDH}=\widehat{BEC}\ (=90^\circ )$

$\Rightarrow\ \triangle BHD\ \backsim\ \triangle BCE\ (g-g)$

$\Rightarrow\ \dfrac{BH}{BC}=\dfrac{BD}{BE}$

$\Leftrightarrow\ BH.BE=BC.BD\ \ \ \ \ \ \ \color{red}{(4)}$

$\hspace{0.65cm}\ $Từ $(4),(5)\ \Rightarrow\ BH.BE+CH.CF=BC.BD+CB.CD$

$\hspace{7.72cm}\ =BC(BD+CD)$

$\hspace{7.72cm}\ =BC^2\ \ (đpcm)$