# 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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You might have confused that with ratio of areas for similar figures.

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 $1^2:3^2$.

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but if two length are in some proportion we could not consider as similar? – giorgi Aug 11 '13 at 20:51
For triangles, two sides in proportion is not sufficient. What about the remaining side? – peterwhy Aug 11 '13 at 20:55
that is what i wanted,so we can't determine also portions of height right if one one pair of sides portion is given – giorgi Aug 11 '13 at 20:57
For the case in your figure, if you consider the bases of the triangles to be AC and CD, then both heights of the triangles are the distance between the parallel lines. – peterwhy Aug 11 '13 at 21:01
but in general case we can't right – giorgi Aug 11 '13 at 21:02