In a quadrilateral $ABCD, $ it is given that $AB$ is parallel to $CD$... 
In a quadrilateral $ABCD, $ it is given that $AB$ is parallel to $CD$ and the diagonals are perpendicular to each other. Show that 
(i)$AD\cdot BC\ge AB\cdot CD$
(ii)$AD+BC\ge AB+CD$

I have no idea how to proceed.
Any help will be appreciated. 

Let $O$ be the point where $AC$ and $BD$ meets. Let $AO = a, BO = b, CO = c, DO = d$. Then, we know that $$\frac{a}{c}=\frac{b}{d}:=t\implies a = tc,  \ \ b =td$$ for some $t$ because $AC\parallel BD$. The first inequality is equivalent to showing $$(a^2+d^2)(b^2+c^2)\geq(a^2+b^2)(c^2+d^2)$$ which is true because $$Left-Right = (a^2-c^2)(b^2-d^2) = c^2d^2(t^2-1)^2\geq 0\implies Left\geq Right$$
The second inequality is equivalent to showing $$\sqrt{a^2+d^2} + \sqrt{b^2+c^2}\geq\sqrt{a^2+b^2}+\sqrt{c^2+d^2}$$
Taking the square on both sides, it is equivalent to showing $$\sqrt{(a^2+d^2)(b^2+c^2)}\geq\sqrt{(a^2+b^2)(c^2+d^2)}$$
which is the same as the first inequality. 
