portion of areas of two triangle

let us consider following picture

we are given that this two line is parallel and also $AC=1/3 * AD$,we should find portion of areas of $ABC$ and $BCD$,now because $AB$ is one third of $AD$, it means that $AC=x$ and $AD=3*x$,from which $CD=2*x$,now as i know , ratios of areas of two triangle is equal to square of ratios of two sides of this triangles,or in other word,if

$AC/CD=1/2$

ratios of their areas should be $1/4$,but answer is simply $1/2$,because $ABC$ is oblique,we may say that these two triangle has same height,but does it change something?suppose that we have two any triangle with ratios of sides be let say $1:3$,what should be ratios of their areas? thanks in advance

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For this case, since the area of triangle is $\frac{1}{2}$base $\times$ height, another triangle with twice the base would have twice the area, given the same height.
Your last question is different, since now the 2nd triangle has triple side lengths for all three sides, the two triangles are similar, and the ratio of areas is $1^2:3^2$.