giúp mình với ạaaa!!

Giải thích các bước giải:

a.Xét $\Delta ADE,\Delta ABD$ có:
Chung $\hat A$
$\widehat{AED}=\widehat{ADB}(=90^o)$
$\to \Delta ADE\sim\Delta ABD(g.g)$
$\to \dfrac{AD}{AB}=\dfrac{AE}{AD}$
$\to AD^2=AB\cdot AE$
Tương tự chứng minh được: $AD^2=AF\cdot AC$
$\to AB\cdot AE=AD^2=AF\cdot AC$
b.Xét $\Delta AFE,\Delta ABC$ có:
Chung $\hat A$
$\dfrac{AF}{AB}=\dfrac{AE}{AC}$ vì $AE\cdot AB=AF\cdot AC$
$\to \Delta AFE\sim\Delta ABC(c.g.c)$
$\to \widehat{AFE}=\widehat{ABC}$
c.Từ câu b $\to \widehat{AEF}=\widehat{ACB}$
$\to \widehat{IEB}=\widehat{AEF}=\widehat{ACB}=\widehat{FCB}$
$\to \widehat{IED}=\widehat{IEB}+\widehat{BED}=\widehat{IEB}+90^o=\widehat{BCF}+90^o=\widehat{DCF}+90^o=\widehat{FCD}+\widehat{DFC}=\widehat{FDI}$
Mà $\widehat{EID}=\widehat{FIC}$
$\to \Delta IED\sim\Delta IDF(c.g.c)$
$\to \dfrac{IE}{ID}=\dfrac{ID}{IF}$
$\to ID^2=IE\cdot IF$

 d.Vì $H$ là trực tâm $\Delta ABC$

Gọi $BH\cap AC=G\to BH\perp AC$

$\to \widehat{BHD}=90^o-\widehat{HBD}=90^o-\widehat{GBC}=\widehat{GCB}=\widehat{ACD}=\widehat{FCD}$

Xét $\Delta IDE,\Delta IDF$ có:

Chung $\hat I$

$\dfrac{ID}{IE}=\dfrac{IF}{ID}$ vì $ID^2=IE\cdot IF$

$\to \Delta IED\sim\Delta IDF(g.g)$

$\to \widehat{IED}=\widehat{IDF}=180^o-\widehat{FDC}=90^o+90^o-\widehat{FDC}=90^o+\widehat{FCD}=90^o+\widehat{BHD}=\widehat{IBK}$

Vì $\widehat{EID}=\widehat{KIB}$

$\to \Delta IED\sim\Delta IBK(c.g.c)$

$\to \dfrac{IE}{IB}=\dfrac{ID}{IK}$

$\to \dfrac{IE}{ID}=\dfrac{IB}{IK}$

Mà $\widehat{EIB}=\widehat{KID}$

$\to \Delta IBE\sim\Delta IKD(c.g.c)$

$\to \widehat{IDK}=\widehat{IEB}=\widehat{ACB}$

$\to DK//AC$

Do $BK\perp AC\to DK\perp BH$