a)

Xét $\Delta ABG$ và $\Delta ACE$, ta có:

+ $\widehat{BAG}$ là góc chung

+ $\widehat{AGB}=\widehat{AEC}=90{}^\circ $

Nên $\Delta ABG\backsim\Delta ACE\left( g.g \right)$

Xét $\Delta ADH$ và $\Delta ACF$, ta có:

+ $\widehat{DAH}$ là góc chung

+ $\widehat{AHD}=\widehat{AFC}=90{}^\circ $

Nên $\Delta ADH\backsim\Delta ACF\left( g.g \right)$

Ta có:

+ $CE\bot AB\,,\,AB//CD\,\Rightarrow CE\bot CD\Rightarrow \widehat{ECB}+\widehat{BCD}=90{}^\circ $

+ $CF\bot AD\,,\,AD//BC\,\Rightarrow CF\bot BC\Rightarrow \widehat{FCD}+\widehat{BCD}=90{}^\circ $

$\Rightarrow \widehat{ECB}=\widehat{FCD}$

Xét $\Delta BEC$ và $\Delta DFC$, ta có:

+ $\widehat{ECB}=\widehat{FCD}\left( cmt \right)$

+ $\widehat{BEC}=\widehat{DFC}=90{}^\circ $

Nên $\Delta BEC\backsim\Delta DFC\left( g.g \right)$

b)

Vì $\Delta ABG\backsim\Delta ACE\left( cmt \right)$

$\Rightarrow \dfrac{AB}{AC}=\dfrac{AG}{AE}\Rightarrow AB.AE=AG.AC$

c)

Vì $\Delta ADH\backsim\Delta ACF\left( g.g \right)$

$\Rightarrow \dfrac{AD}{AC}=\dfrac{AH}{AF}\Rightarrow AD.AF=AH.AC$

Xét $\Delta AHD$ và $\Delta CGB$, ta có:

+ $\widehat{AHD}=\widehat{CGB}=90{}^\circ $

+ $AD=CB$ (vì $ABCD$ là hình bình hành)

+ $\widehat{HAD}=\widehat{GCB}$ (vì $AD//BC$, hai góc so le trong)

Nên $\Delta AHD=\Delta CGB\left( ch-gn \right)$

Do đó $AH=CG$

Ta có $\begin{cases}AB.AE=AG.AC\\AD.AF=AH.AC\end{cases}\,\,\,\left(cmt\right)$

$\Rightarrow AB.AE+AD.AF=AC\left( AG+AH \right)$

$\Rightarrow AB.AE+AD.AF=AC\left( AG+CG \right)$

$\Rightarrow AB.AE+AD.AF=AC.AC=A{{C}^{2}}$