Giải thích các bước giải:
a.Xét $\Delta AMK,\Delta DMC$ có:
$\widehat{KAM}=\widehat{MDC}(=90^o)$
$\widehat{AMK}=\widehat{DMC}$
$\to \Delta AMK\sim\Delta DMC(g.g)$
b.Xét $\Delta CDM, \Delta CAB$ có:
Chung $\hat C$
$\widehat{CDM}=\widehat{CAB}(=90^o)$
$\to \Delta CDM\sim\Delta CAB(g.g)$
$\to \dfrac{CD}{CA}=\dfrac{CM}{CB}$
$\to CD.CB=CM.CA$
c.Xét $\Delta BDK, \Delta BAC$ có:
chung $\hat B$
$\widehat{BAC}=\widehat{BDK}(=90^o)$
$\to \Delta BAC\sim\Delta BDK(g.g)$
$\to \dfrac{BA}{BD}=\dfrac{BC}{BK}$
$\to BA.BK=BD.BC$
d.Từ câu c
$\to \dfrac{BA}{BC}=\dfrac{BD}{BK}$
Mà $\widehat{ABD}=\widehat{KBC}$
$\to \Delta BAD\sim\Delta BCK(c.g.c)$
$\to \widehat{BDA}=\widehat{BKC}$
e.Ta có $\widehat{ABC}=60^o, \hat A=90^o$
$\to \Delta ABC$ là nửa tam giác đều
$\to \dfrac{AB}{BC}=\dfrac12$
Từ câu d
$\to \dfrac{S_{BAD}}{S_{BCK}}=(\dfrac{BA}{BC})^2=\dfrac14$
$\to S_{BAD}=\dfrac14S_{BCK}=39(cm^2)$