a: Xét ΔABH vuông tại H có cos B=BH/AB
=>AB=2a
=>BC=4a
=>\(AC=2\sqrt{3}a\)
b: BH*BC=BA^2
BD*BK=BA^2
DO đó; BH*BC=BD*BK
=>BH/BK=BD/BC
=>ΔBHD đồng dạng với ΔBKC
=>\(\dfrac{S_{BHD}}{S_{BKC}}=\left(\dfrac{BH}{BK}\right)^2=\left(\dfrac{1}{2}AB:BK\right)^2=\dfrac{1}{4}\cdot\left(\dfrac{AB}{BK}\right)^2\)
\(=\dfrac{1}{4}\cdot cos^2ABD\)
=>\(S_{BHD}=\dfrac{1}{4}\cdot cos^2ABD\cdot S_{BKC}\)