Giải thích các bước giải:

a.Ta có:

$CD\perp BC\to BD^2=BC^2+CD^2\to CD^2=\sqrt{BD^2-BC^2}=9$

$\to\sin\widehat{BDC}=\dfrac{BC}{BD}=\dfrac45$

Ta có : $\widehat{DCB}=\widehat{DBE}=90^o,\widehat{BDC}=\widehat{BDE}$

$\to\Delta DBC\sim\Delta DEB(g.g)$

$\to\dfrac{BD}{DE}=\dfrac{DC}{DB}$

$\to DE=\dfrac{DB^2}{DC}=25$

$\to CE=DE- CD=16$

$\to BE=\sqrt{BC^2+CE^2}=20$

Ta có $\widehat{ABC}=\widehat{ACB}$

$\to 90^o-\widehat{ABC}=90^o-\widehat{ACB}$

 $\to \widehat{AEC}=\widehat{ACE}$

$\to AE=AC$

Mà $AB=AC\to AB=AE\to A$ là trung diểm $BE\to AB=\dfrac12BE=10$

b.Ta có: $\widehat{BFD}=\widehat{DBA}=90^o,\widehat{BDF}=\widehat{BDA}$

$\to \Delta DBF\sim\Delta DAB(g.g)$

$\to \dfrac{DB}{DA}=\dfrac{DF}{DB}$

$\to DB^2=DF.DA$

$\to DF.DA=DC.DE(=DB^2)$

Lại có $\widehat{AFB}=\widehat{ABD}=90^o,\widehat{BAF}=\widehat{BAD}$

$\to \Delta ABF\sim\Delta ADB(g.g)$

$\to\dfrac{AB}{AD}=\dfrac{AF}{AB}$

$\to\dfrac{AC}{AD}=\dfrac{AF}{AC}$ vì $AB=AC$
Mà $\widehat{FAC}=\widehat{DAC}$

$\to \Delta ACF\sim\Delta ADC(c.g.c)$

$\to \widehat{ACF}=\widehat{ADC}$

c.Ta có $DH\perp BD\to DH//BA$

$\to \dfrac{IH}{IA}=\dfrac{DH}{AB}=\dfrac{DH}{AE}=\dfrac{HC}{AC}$

$\to \dfrac{IA}{IH}=\dfrac{AC}{HC}$

$\to \dfrac{HI+HA}{IH}=\dfrac{HA-HC}{HC}$

$\to 1+\dfrac{HA}{HI}=\dfrac{HA}{HC}-1$

$\to \dfrac{HA}{HC}-\dfrac{HA}{HI}=2$

$\to\dfrac{1}{HC}-\dfrac{1}{HI}=\dfrac{2}{HA}$

d.Ta có $AK=AB\to AK=AC$

$\to \widehat{AKC}=\widehat{ACK}$

$\to \widehat{KCD}+\widehat{KDC}=\widehat{ACF}+\widehat{FCK}$

$\to \widehat{KCD}+\widehat{ADC}=\widehat{ACF}+\widehat{FCK}$

$\to \widehat{KCD}=\widehat{ACF}$

$\to CK$ là phân giác $\widehat{FCD}$