Let $ABCD$ be a cyclic quadrilateral such that $AB \cdot BC=2 \cdot AD \cdot DC$. Prove that its diagonals $AC$ and $BD$ satisfy the inequality $8BD^2 \le 9AC^2$.
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Is this correct? Seems too easy. The given condition is $\text{Area}(ABC)=2\text{Area}(ACD)$. Let $AC\cap BD=E$ then from the given condition we see that $\Delta{DH_1E}\sim\Delta{BH_2E}$ where $H_1$ is the foot of perpendicular from $D$ to $AC$ and $H_2$ is the foot of perpendicular from $B$ to $AC$. This gives $BE=2ED$. Then we have :
\[ AC^2=(AE+EC)^2\ge 4AE\cdot EC=4BE\cdot ED=\frac{8}{9}BD^2 \\ \implies 9AC^2\ge 8BD^2 \]