Giải thích các bước giải:

a.Xét $\Delta ADE, \Delta DCF$ có:

$AE=DF$

$\widehat{EAD}=\widehat{FDC}(=90^o)$

$AD=DC$

$\to \Delta ADE=\Delta DCF(c.g.c)$

$\to \widehat{AED}=\widehat{CFD}$

$\to \widehat{AED}=\widehat{DFH}$

$\to \Delta DHF\sim\Delta DEA(g.g)$

$\to \widehat{DHF}=\widehat{DAE}=90^o$

$\to DE\perp CF$

c.Xét $\Delta DHF,\Delta CFD$ có:

Chung $\hat F$

$\widehat{FHD}=\widehat{FDC}(=90^o)$

$\to \Delta DFH\sim\Delta CFD(g.g)$

$\to \dfrac{DF}{CF}=\dfrac{FH}{DF}$

$\to DF^2=FH\cdot FC$

Mà $AE=DF$

$\to AE^2=HF\cdot FC$

c.Gọi $DE\cap AC=G$

Ta có:

$\widehat{EKC}$

$=180^o-\widehat{KEC}-\widehat{KCE}$

$=180^o-(\widehat{KED}+\widehat{DEC})-(\widehat{KCF}+\widehat{FCE})$

$=180^o-(\widehat{KED}+\widehat{KCF})-(\widehat{DEC}+\widehat{ECF})$

$=180^o-(\dfrac12\widehat{AED}+\dfrac12\widehat{DCF})-(\widehat{HEC}+\widehat{HCE})$

$=180^o-\dfrac12(\widehat{AED}+\widehat{DCF})-90^o$

$=90^o-\dfrac12(\widehat{AED}+\widehat{ADE})$

$=90^o-\dfrac12\cdot 90^o$

$=45^o$