Sửa đề: $OK\perp SA$
Hình thoi $ABCD$ cạnh $a$ có:
$\widehat{BAD}= 60^\circ\quad (gt)$
$\to ∆ABD$ đều
$\to \begin{cases}AB = BC = CD = DA = BD = a\\OA = OC =\dfrac{a\sqrt3}{2}\\OB = OD =\dfrac a2\end{cases}$
Xét $∆OKA$ và $∆SCA$ có:
$\widehat{K}=\widehat{C}=90^\circ$
$\widehat{A}:$ góc chung
Do đó $∆OKA\sim ∆SCA\, (g.g)$
$\to \dfrac{OK}{SC}=\dfrac{OA}{SA}$
$\to OK =\dfrac{SC.OA}{SA}$
$\to OK =\dfrac{SC.OA}{\sqrt{SC^2 + AC^2}}$
$\to OK =\dfrac{\dfrac{a\sqrt6}{2}\cdot\dfrac{a\sqrt3}{2}}{\sqrt{\dfrac{3a^2}{2}+3a^2}}=\dfrac a2$
$\to OK = OB = OD =\dfrac a2$
$\to ∆BKD$ vuông tại $K$
$\to \widehat{BKD}=90^\circ$
$\to DK\perp BK\qquad (1)$
Ta có:
$BD\perp AC$ (hình thoi)
$BD\perp SC\quad (SC\perp (ABCD))$
$\to BD\perp (SAC)$
$\to BD\perp SA$
Lại có: $OK\perp SA\quad (gt)$
$\to SA\perp (BKD)$
$\to SA\perp BK\qquad (2)$
Từ $(1)(2)\to BK\perp (SAD)$
mà $BK\subset (SAB)$
nên $(SAB)\perp (SAD)$
