Đáp án:

Ta có $AH\,\bot\,BC\Rightarrow\widehat{AFB}+\widehat{HBF}=90^\circ$

$CE\,\bot\,BF\Rightarrow\widehat{BCE}+\widehat{HBF}=90^\circ$

$\Rightarrow\widehat{AFB}=90^\circ-\widehat{HBF}=\widehat{BCE}$

Ta có $\Delta BDE\backsim\Delta CDA$ (góc-góc)

$\Rightarrow\dfrac{DE}{DA}=\dfrac{DB}{DC}$ mà có hai góc đối đỉnh

$\Rightarrow\Delta ADE\backsim\Delta CDB$ (cạnh-góc-cạnh)

$\Rightarrow\widehat{BCE}=\widehat{ABE}$ mà $\widehat{AFB}=\widehat{BCE}$

$\Rightarrow\widehat{ABE}=\widehat{AFB}$ mà có chung góc tại $B$

$\Rightarrow\Delta ABE\backsim\Delta FBA$ (góc-góc)

$\Rightarrow\dfrac{AB}{BE}=\dfrac{BF}{AB}\Rightarrow AB^2=BE.BF$

Mà $BG=AB\Rightarrow BG^2=BE.BF$

$\Rightarrow\dfrac{BE}{BG}=\dfrac{BG}{BF}$ mà có góc chung tại $B$

$\Rightarrow\Delta BEG\backsim\Delta BGF$ (cạnh-góc-cạnh)

$\Rightarrow\widehat{BGF}=\widehat{BEG}=90^\circ\Rightarrow BG\,\bot\,FG$