Giải thích các bước giải:

a.Xét $\Delta ABI,\Delta ACE$ có:

Chung $\hat A$

$\hat I=\hat E(=90^o)$

$\to \Delta ABI\sim\Delta ACE(g.g)$

$\to\dfrac{AB}{AC}=\dfrac{AI}{AE}$

$\to AB.AE=AI.AC$

b.Xét $\Delta CBI,\Delta ACF$ có:

$\hat I=\hat F(=90^o)$

$\widehat{ICB}=\widehat{CAF}$

$\to \Delta CBI\sim\Delta ACF(g.g)$

$\to \dfrac{CI}{AF}=\dfrac{BC}{AC}$

$\to AF.BC=CI.AC$

$\to AB.AE+AF.BC=AI.AC+CI.AC=AC^2$

c.Ta có: $CE\perp AB, CF\perp AD$

               $ABCD$ là hình bình hành

$\to S_{ABC}=S_{ACD}$

$\to \dfrac12CE.AB=\dfrac12CF.AD$

$\to CE.AB=CF.AD$

$\to \dfrac{CE}{AD}=\dfrac{CF}{AB}$

$\to \dfrac{CE}{BC}=\dfrac{CF}{AB}$

Mà $\widehat{ECF}=360^o-\hat E-\hat F-\widehat{EAF}=180^o-\widehat{BAD}=\widehat{ABC}$

$\to \Delta CEF\sim\Delta BCA(g.g)$

$\to \widehat{CEF}=\widehat{BCA}$