Giải thích các bước giải:

a.Xét $\Delta AEB,\Delta CEA$ có:

$\widehat{ABE}=\widehat{EAC}(=90^o)$

Chung $\hat E$

$\to\Delta AEB\sim\Delta CEA(g.g)$

Ta có:$EC^2=AC^2+AE^2\to AE^2=CE^2-AC^2=256$

$\to EC=16$

Do $AB\perp CE$

$\to AB\cdot CE=AE\cdot AC(=2S_{ABC})$

$\to AB=\dfrac{48}5$

b.Ta có $BF//AC(\perp AE)$

$\to \dfrac{BF}{AC}=\dfrac{EF}{EA}$

$\to \dfrac{BF}{EF}=\dfrac{AC}{EA}$

Tương tự câu a $\to\Delta CAB\sim\Delta CEA(g.g)$

$\to \dfrac{AB}{EA}=\dfrac{CB}{CA}$

$\to \dfrac{AC}{EA}=\dfrac{CB}{AB}$

$\to \dfrac{BF}{EF}=\dfrac{CB}{AB}$

$\to BF\cdot AB=BC\cdot EF$

c.Ta có $ABCD$ là hình chữ nhật

$\to AC=BD, AD=BC, AB=CD$

Vì $BF//AC$

$\to \dfrac{BF}{AC}=\dfrac{BE}{EC}$

$\to \dfrac{BF}{AC-BF}=\dfrac{BE}{EC-BE}$

$\to \dfrac{BF}{AC-BF}=\dfrac{BE}{BC}$

$\to \dfrac{BF}{BD-BF}=\dfrac{BE}{AD}$
$\to \dfrac{AD}{BD-BF}=\dfrac{BE}{BF}$

Xét $\Delta BFE,\Delta ABC$ có:

$\widehat{BFE}=\widehat{ABC}(=90^o)$

$\widehat{FBE}=\widehat{ACB}$ vì $FB//AC$

$\to\Delta BEF\sim\Delta CAB(g.g)$

$\to \dfrac{BF}{CB}=\dfrac{BE}{CA}$

$\to \dfrac{BE}{BF}=\dfrac{CA}{CB}=\dfrac{AC}{AD}$

$\to \dfrac{AD}{BD-BF}=\dfrac{AC}{AD}$

$\to AD^2=AC(BD-BF)$

d.Gọi $CF\cap AD=H, EM\cap AD=G$

Ta có: $AD//EC$

$\to\dfrac{AH}{CE}=\dfrac{FA}{FE}=\dfrac{BC}{BE}$

$\to \dfrac{AH}{BC}=\dfrac{CE}{BE}$

Lại có:

$\dfrac{KA}{KB}=\dfrac{AG}{BE}$

Do $ABCD$ là hình chữ nhật, $AC\cap BD=M$

$\to M$ là trung điểm $AC, DB$

$\to\dfrac{AG}{CE}=\dfrac{MA}{MC}=1$

$\to AG=CE$

$\to \dfrac{AH}{BC}=\dfrac{CE}{BE}=\dfrac{AG}{BE}=\dfrac{KA}{KB}$

Lại có $\widehat{HAK}=\widehat{KBC}(=90^o)$

$\to\Delta AHK\sim\Delta BCK(c.g.c)$

$\to \widehat{AKH}=\widehat{BKC}$

$\to C, K, F$ thẳng hàng