Giải thích các bước giải:
a.Xét $\Delta BHK,\Delta BDC$ có:
Chung $\hat B$
$\widehat{BKH}=\widehat{BDC}(=90^o)$
$\to\Delta BKH\sim\Delta BDC(g.g)$
b.Xét $\Delta CHK,\Delta CBE$ có:
Chung $\hat C$
$\widehat{CKH}=\widehat{CEB}(=90^o)$
$\to\Delta CHK\sim\Delta CBE(g.g)$
$\to \dfrac{CH}{CB}=\dfrac{CK}{CE}$
$\to CH.CE=CK.CB$
c.Từ câu a $\to \dfrac{BH}{BC}=\dfrac{BK}{BD}$
$\to BH.BD=BK.BC$
$\to BH.BD+CH.CE=BK.BC+CK.CB=BC^2$
d.Ta có $\hat A=60^o, \widehat{AEC}=90^o\to \Delta ACE$ là nửa tam giác đều
$\to AE=\dfrac12AC$
$\to \dfrac{AE}{AC}=\dfrac12$
Xét $\Delta ACE,\Delta ABD$ có:
Chung $\hat A$
$\widehat{AEC}=\widehat{ADB}(=90^o)$
$\to\Delta ABD\sim\Delta ACE(g.g)$
$\to \dfrac{AB}{AC}=\dfrac{AD}{AE}$
$\to\dfrac{AB}{AD}=\dfrac{AC}{AE}$
Mà $\widehat{BAC}=\widehat{EAD}$
$\to\Delta ABC\sim\Delta ADE(c.g.c)$
$\to \dfrac{S_{ADE}}{S_{ABC}}=(\dfrac{AE}{AC})^2=\dfrac14$
$\to S_{ADE}=\dfrac14S_{ABC}=30(cm^2)$
