Giải thích các bước giải:

a.Ta có: $DD'//BC$

$\to \widehat{BDD'}=180^o-\widehat{DBC}=\widehat{DD'C}$

$\to BDD'C$ là hình thang cân

$\to DC=D'B, BD=CD'$

$\to \widehat{DAC}=\widehat{BAD'},\widehat{DAB}=\widehat{CAD'}$

$\to\widehat{DAC}=\widehat{BAE},\widehat{DAB}=\widehat{CAE}$

Xét $\Delta ABD,\Delta ACE$ có:

$\widehat{DAB}=\widehat{CAE}$

$\widehat{ADB}=180^o-\widehat{ACB}=\widehat{ACE}$

$\to\Delta ABD\sim\Delta AEC(g.g)$

Xét $\Delta ADC,\Delta ABE$ có:

$\widehat{ADC}=\widehat{ABE}$

$\widehat{DAC}=\widehat{BAE}$

$\to\Delta ADC\sim\Delta ABE(g.g)$

b.Từ câu a ta có: $\Delta ADC\sim\Delta ABE$

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

$\to AD\cdot AE=AB\cdot AC$

c.Xét $\Delta AFD,\Delta AD'B$ có:

$\widehat{DAF}=\widehat{DAB}+\widehat{BAC}=\widehat{D'AC}+\widehat{BAC}=\widehat{BAD'}$

$\widehat{ADF}=\widehat{ADD'}=\widehat{ABD'}$

$\to\Delta AFD\sim\Delta AD'B(g.g)$