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I was wondering is it possible to solve this without assuming that CAD=DAB. As I use the law of sines, trigonometry and have tried to apply law of cosines. However, I cannot see how you can solve this with just using a and $\alpha$.

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    sorry, i didn't see figure properly. I guess the answer is no, as we can change $AD$ freely within $\alpha $ and yet construct a right angled triangle to fit it2012-12-26
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    Extend $CA$ a little, to $CA'$. Extend $CB$ a little, to $CB'$, so that $A'B'$ is parallel to $AB$. Then we can find $D'$ in $AD$ such that $A'D'=AD=a$. The shaded area has increased. Your picture was very helpful.2012-12-26
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    Yes indeed, the question was well expressed.2012-12-26

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