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In traingle ABC, Angle A=45 degrees, Angle B is 60 degrees, and AC= radical 15. D is also a point on AB so that AB is perpendicular to CD. The circle with diameter AB intersects CD at point E. Compute (DE)^2.

So this is what I have done so far, since I just joined today and I never knew I had to post what work I done so far. Since we know A and B, we know that C is 75 degrees. Since AB is perpendicular to DC we know that triangle ADC is isosceles and traingle DBC is a 30, 60, 90 triangle. I could find the lengths of the sides of triangle ADC and triangle ADC but how would I find DE?

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    Aren't we missing some info? The problem description doesn't seem to pin down point $B$ (though there are restrictions). Yet, the longer $AB$ is, the longer $DE$ is, so that we have no unique solution. If $B$ and $D$ coincide, then $E$ coincides with them, making $DE = 0$. If $B$ is even a hair further away from $A$, then $DE$ is non-zero. The other extreme is when the circle is large enough so that $E$ matches $C$, in which case $\angle ACB$ is a right angle (why?) and $AB =$ ___. In any (non-zero) case, $\angle AEB$ is a right angle (same "why?"), so that $AD/DE=ED/DB$ (different "why?").2011-10-27
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    BTW, I agree with what has been noted in a comment to another of your (non-)questions: unmotivated contest problems aren't appropriate for this site (though I do appreciate that you've made an effort to note the source). I chose to "answer" above because I found the problem, as posed, unsolvable, and found that to be interesting. :)2011-10-27
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    Believe me I have tried to answer this problem. I have recieved the answer after the competition was completed. But it still seems unsolvable to me at this point >_<2011-10-27
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    Ah ... You've edited to include the measure of angle $B$! Okay, then: Now, you can get the length of $DB$. (Use the Law of Sines to get $AB$ and go from there, or else use the geometry of the 45-45 triangle $ADC$ and 30-60 triangle $BDC$. (The latter is probably what the problem intends.)) Then, you can use the proportion at the end of my first comment. (The proportion follows from that fact that $\angle AEB$ is a right angle; as is *any* angle (as they say) "inscribed in a semi-circle".)2011-10-27
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    Thank you! I am not used to using proportions. But I really appreciate the help.2011-10-27

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