Area of Triangle ABC Given Dimensions

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In triangle ABC $AC=BD, CE=2, ED=1, AE=4$ and $\angle CAE=2 \angle DAB$. Find area ABC.
 
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  • Triangle ABC with points E on side CB... wait, E lies on segment AC-side and D on side BC; angles at A marked 2α between AC and AE and α between AD and AB, with points C, E, D, B labelled.
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Beer induced request follows.
maxkor said:
In triangle ABC $AC=BD, CE=2, ED=1, AE=4$ and $\angle CAE=2 \angle DAB$. Find area ABC.
Please show us what you have tried and exactly where you are stuck.
We can't help you if we don't where you are stuck.
 
This is what it looks like, but how to justify the red ones or maybe there is another way?
 
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  • Geometry diagram of triangle ABC with B on a horizontal line at left, D and E on that line, and C on it near right. Lines run from A to B, D, E and C. Labels show BD = 3, DE = 1, EC = 2, AE = 4, AC = 3, C to foot of perpendicular = 3/4, angles alpha, 2alpha, and red dashed altitude from A.
    154882.png
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In your diagram, all three sides of $\triangle AEC$ are known. Using the law of cosines you get $2\alpha = \cos^{-1}(7/8)$ and $\alpha \approx 14.47751219^{\circ}$.

Continuing to use the law of cosines we get $AD=\dfrac{3\sqrt{10}}{2}$ and $AB=3\sqrt{6}$.

Finally, using the law of cosines on $\triangle ABD$ we get $\alpha \approx 71.170769^{\circ}$.

Your diagram is incorrect.