In triangle ABC $AC=BD, CE=2, ED=1, AE=4$ and $\angle CAE=2 \angle DAB$. Find area ABC.
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#3
maxkor
90
0
This is what it looks like, but how to justify the red ones or maybe there is another way?
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#4
HOI
923
2
How did you get those?
#5
maxkor
90
0
Geogebra...
#6
mrtwhs
48
0
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 $\alpha \approx 71.170769^{\circ}$.