I've attempted to mark segment $AO$ as $a$ and $CO$ as $b$.
Now, if draw a line from $D$ that is perpendicular to $AC$, and draw another line from B that is perpendicular to $AC$, then we can use some variables to express the area of $\triangle BOC$ and $\triangle AOD$.
However, when I try to make a equation, I found that I have too many variables.
How can I move further?
On
Altitude from $B$ to $AC$, $h_1$ is $\dfrac{2[AOB]}{AO}=\dfrac{2\triangle_{1}}{a}$
Altitude from $D$ to $AC$, $h_2$ is $\dfrac{2[COD]}{CO}=\dfrac{2\triangle_{2}}{b}$
So $$[ABCD] = [AOB]+[COD]+[BOC]+[DOA]$$
$$ = \triangle_{1}+\triangle_{2}+\dfrac{1}{2}\cdot b \cdot \dfrac{2\triangle_{1}}{a}+\dfrac{1}{2}\cdot a \cdot\dfrac{2\triangle_{2}}{b}$$
$$ = \triangle_{1}+\triangle_{2}+ \dfrac{b}{a}\triangle_{1}+ \dfrac{a}{b}\triangle_{2}$$
which is minimum when (their product is constant so their sum is minimum when both equal)
$$\dfrac{b}{a}\triangle_{1} = \dfrac{a}{b}\triangle_{2}$$ $$\Rightarrow \dfrac{\triangle_{1}}{\triangle_{2}} = \dfrac{a^2}{b^2}$$
Obtain same expression by taking altitudes on other diagonal.
Since $\angle AOB = \angle COD$, we conclude $[ABCD]$ is minimum when $$\triangle AOB \sim \triangle COD$$
As a result $\angle BAO = \angle DCO \Rightarrow AB || CD$
Thus $ABCD$ is a trapezium.
You can substitute back to find the minimum value is
$$\bbox[#FFFF33,5px]{[ABCD]_{\text{min}} = (\sqrt{\triangle_{1}}+\sqrt{\triangle_{2}})^2}$$
We need to use the fact that the ratio of triangles with the same height is equal to the ratio of their bases.
Suppose $AO : OC = 1: k$. Then $[\triangle AOD] : [\triangle COD] = AO:OC = [\triangle AOB] : [\triangle COB] = 1:k$.
This gives $[\triangle AOD] = \frac4k$ and $[\triangle COB] = 9k$.
Now the area of the quadrilateral is equal to
$$4 + 9+\frac4k+9k=4+9+12+\frac4k-12+9k=25+\left(\frac{2}{\sqrt k}-3\sqrt{k}\right)^2 \ge 25$$
Equality holds when $k = \frac23$, so the minimum area is $25$.
We can use $AM \ge GM: a+b \ge 2\sqrt {ab}$ to obtain $\dfrac 4k + 9k \ge 12$ as well.