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:

$\dfrac{S_{BHD}}{S_{BKC}}=\left(\dfrac{1}{2}\cos\widehat{ABD}\right)^2$

$\Rightarrow \dfrac{S_{BHD}}{S_{BKC}} = \dfrac{1}{4}\cos^2\widehat{ABD}$

$\Rightarrow S_{BHD} = \dfrac{1}{4}S_{BKC}\cos^2\widehat{ABD}$