Ta có: SBCD = \(\dfrac{1}{2}\)DH.BC = \(\dfrac{1}{2}\).12.DH = 6DH
SABH = \(\dfrac{1}{2}\)AH.BH = \(\dfrac{1}{2}\).4.AH = 2AH
Vì AD = 2DH
\(\Rightarrow\) AD + DH = 2DH + DH
\(\Rightarrow\) AH = 3DH (D \(\in\) AH)
\(\Rightarrow\) DH = \(\dfrac{1}{3}\)AH
\(\Rightarrow\) 6DH = 2AH
\(\Rightarrow\) SBCD = SABH
Chúc bn học tốt!