Giải thích các bước giải:

a.Xét $\Delta ADE, \Delta ECD$ có:

$\widehat{DAB}=\widehat{DEC}(=90^o)$

$\widehat{ADB}=\widehat{EDC}$(đối đỉnh)

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

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

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

Mà $\widehat{ADE}=\widehat{BDC}$

$\to \Delta ADE\sim\Delta BDC(c.g.c)$

b.Từ câu a $\to \widehat{DBA}=\widehat{DCE}$

Ta có: $BE\perp MC, CA\perp AB\to CA\perp BM, BE\cap CA=D$

$\to D$ là trực tâm $\Delta MBC\to MD\perp BC$

Do $DK\perp BC$

$\to M, D, K$ thẳng hàng

Xét $\Delta MDE,\Delta MCK$ có:

Chung $\hat M$

$\widehat{MED}=\widehat{MKC}(=90^o)$

$\to \Delta MDE\sim\Delta MCK(g.g)$

$\to \dfrac{MD}{MC}=\dfrac{ME}{MK}$

$\to \dfrac{MD}{ME}=\dfrac{MC}{MK}$

Do $\widehat{KME}=\widehat{DMC}$

$\to \Delta MKE\sim\Delta MCD(c.g.c)$

$\to \widehat{MKE}=\widehat{MCD}$

$\to \widehat{DKI}=\widehat{DCE}$

Tương tự chứng minh được $\widehat{DKA}=\widehat{DBA}$

Do $\widehat{DBA}=\widehat{DCE}$

$\to \widehat{DKI}=\widehat{DKA}$

$\to KD$ là phân giác $\widehat{AKI}$

Mà $KD\perp KC\to KC$ là phân giác ngoài tại đỉnh $K$ của $\Delta KAI$

$\to \dfrac{CA}{CI}=\dfrac{KA}{KI}=\dfrac{DA}{DI}$

$\to ID\cdot CA=DA\cdot CI$

c.Trên $AC$ lấy $F$ sao cho $\widehat{AEF}=\widehat{BEC}$

Xét $\Delta EAF,\Delta EBC$ có:

$\widehat{AEF}=\widehat{BEC}$

$\widehat{EAF}=\widehat{DBC}=\widehat{EBC}$

$\to \Delta EAF\sim\Delta EBC(g.g)$

$\to \dfrac{EA}{EB}=\dfrac{AF}{BC}$

$\to AE\cdot BC=BE\cdot AF$

Xét $\Delta EAB,\Delta EFC$ có:

$\widehat{ABE}=\widehat{ABD}=\widehat{ECD}=\widehat{ECF}$

$\widehat{AEB}=\widehat{AEF}-\widehat{BEF}=\widehat{BEC}-\widehat{BEF}=\widehat{FEC}$

$\to \Delta EAB\sim\Delta EFC(g.g)$

$\to \dfrac{AB}{FC}=\dfrac{BE}{CE}$

$\to AB\cdot CE=FC\cdot BE$

$\to AB\cdot CE+AE\cdot BC=FC\cdot BE+BE\cdot AF=BE\cdot (CF+FA)=BE\cdot AC$