Ta có:
$AD\perp BK$
$\Rightarrow \widehat{ADB} = 90^o$
$\Rightarrow \widehat{ADB} = \widehat{AHB} =90^o$
$\Rightarrow ABHD$ là tứ giác nội tiếp
$\Rightarrow \widehat{HAB} = \widehat{HDB}$
mà $\widehat{HAB} = \widehat{ACB}=\widehat{KCB}$ (cùng phụ $\widehat{HAC}$)
nên $\widehat{HDB} = \widehat{KCB}$
Xét $∆BHD$ và $∆BKC$ có:
$\widehat{B}:$ góc chung
$\widehat{HDB} = \widehat{KCB}$
Do đó: $∆BHD\sim ∆BKC\, (g.c)$
$\Rightarrow \dfrac{S_{BHD}}{S_{BKC}}=\left(\dfrac{BH}{BK}\right)^2$
Ta lại có:
$\dfrac{BH}{BK}$
$= \dfrac{\dfrac{AB^2}{BC}}{BK}$
$= \dfrac{AB}{BC}\cdot\dfrac{AB}{BK}$
$= \dfrac{\sqrt{BH.BC}}{BC}.\cos\widehat{ABK}$
$= \dfrac{\sqrt{2.8}}{8}.\cos\widehat{ABD}$
$= \dfrac{1}{2}\cos\widehat{ABD}$
Do đó:
$\dfrac{S_{BHD}}{S_{BKC}}=\left(\dfrac{1}{2}\cos\widehat{ABD}\right)^2$
$\Rightarrow \dfrac{S_{BHD}}{S_{BKC}} = \dfrac{1}{4}\cos^2\widehat{ABD}$
$\Rightarrow S_{BHD} = \dfrac{1}{4}S_{BKC}\cos^2\widehat{ABD}$
