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 $\widehat{FCD}$
