Let $abcd$ be a parallelogram. extend $\overline{bc}$ past $b$ to $f$, and let $e$ be the intersection of $\overline{ab}$ and $\overline{df}$. if the areas of triangles $bef$ and $ade$ are 1 and 9, respectively, find the area of parallelogram $abcd$.

Let's look at two triangles BEF and ADE, these two triangles are similar. By 3 angles. Two angles are vertical and rest angles are equal like a opposite interior angles between two parallels lines (here it's BC and AD by definition of parallelogram). We know that ratio between areas of similars triangles is equal to  the coefficient of similarity in in the square.k^2=1/9k=1/3.That's mean, that DE/EF = 3:1.At another side, we can look at two another triangles: DFC and  BEF.They are similars too, because of BE is parallel to DC - foot of DCF. We remember that DE/EF = 3:1, that's mean if let EF=x, than DE=3x (and DF=4x).We can say that the coefficient of similarity between DFC and  BEF is 1 to 4.And the ratio os areas is 1 to 16. Area of big triangle (DFC) is 16, and area of small (BEF) is 1. Area of BCDE=Area of DFC - area of BEF. 16-1=15   Area of BCDENow it's clearly that For find area of ABCD, we need take area ADE and add  to area BCDE9+15=27The answer is 27.  

Iagree that is the answer to the question

Lol idk ask google. ejie