Giải thích các bước giải:
a.$\widehat{ABC}=90^o-\widehat{ACB}=90^o-36^o=64^o$
b.Ta có:
$\begin{cases}\widehat{ABD}=\widehat{DBE}\\ chung \quad BD\\ \widehat{BAD}=\widehat{BED}(=90^o)\end{cases}\rightarrow\Delta BDA=\Delta DBE(g.c.g)$
c.Ta có:
$\begin{cases}\widehat{KAB}=\widehat{ABD}\\ chung \quad AB\\ \widehat{KBA}=\widehat{BAD}\end{cases}$
$\rightarrow \Delta KAB=\Delta DBA(g.c.g)$
$\rightarrow KA=BD$
d.Vì $BD\perp CF, CA\perp BF\rightarrow D$ là trực tâm $\Delta FBC\rightarrow FD\perp BC$
Mà $\widehat{DEB}=\widehat{DAB}=90^o\rightarrow DE\perp BC$
$\rightarrow D,E,F$ thẳng hàng