cú tuu cú tuiiiii mn

Giải thích các bước giải:

a.Xét $\Delta BCE,\Delta BDE$ có:

Chung $\hat E$

$\widehat{BCE}=\widehat{DBE}(=90^o)$

$\to \Delta BCE\sim\Delta DBE(g.g)$

b.Xét $\Delta CBD,\Delta BCH$ có:

$\widehat{DCB}=\widehat{BHC}(=90^o)$

$\widehat{CBH}=90^o-\widehat{CBD}=\widehat{BDC}$

$\to \Delta CBD\sim\Delta HCB(g.g)$

$\to \dfrac{CB}{HC}=\dfrac{BD}{BC}$

$\to BC^2=CH\cdot BD$

c.Ta có: $ABCD$ là hình bình hành 

$\to CD=AB=8, BC=AD=6$

$\to BD=\sqrt{BC^2+CD^2}=10$

$\to 6^2=CH\cdot 10$

$\to CH=\dfrac{18}5$

Xét $\Delta ECH,\Delta EBD$ có:

Chung $\hat E$

$\widehat{CHE}=\widehat{DBE}(=90^o)$

$\to \Delta ECH\sim\Delta EDB(g.g)$

$\to \dfrac{S_{CEH}}{S_{BED}}=(\dfrac{CH}{BD})^2=\dfrac{81}{625}$