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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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    Thank you! I am not used to using proportions. But I really appreciate the help.2011-10-27

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You found $AD=\sqrt{15}/\sqrt{2}$ and $DB=(\sqrt{15}/\sqrt{2})/\sqrt{3}$. Now the triangle $AEB$ is rectangular, therefore by the "altitude theorem" one has $DE^2=AD\cdot DB=5\sqrt{3}/2$.

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In order to keep the numbers, and more importantly, the typing, simple, define $k$ by $\sqrt{15}=2k\sqrt{6}$. We do some side-chasing, using properties of so-called "special" triangles.

Since $\angle CAD=\angle DCA=45^\circ$, we have $AD=CD=2k\sqrt{3}$. And since $\angle ABC=60^\circ$, the tangent of this angle is $\sqrt{3}$, and therefore $DB=2k$.

Thus $AB=2k(\sqrt{3}+1)$, and therefore the circle has radius $k(\sqrt{3}+1)$. Also, if $O$ is the center of the circle, then $OD=AD-AO=2k\sqrt{3}-k(\sqrt{3}+1)=k(\sqrt{3}-1)$.

By the Pythagorean Theorem for $\triangle ODE$, we have $DE^2=OE^2-OD^2=k^2(\sqrt{3}+1)^2-k^2(\sqrt{3}-1)^2=4\sqrt{3}k^2.$
But $k^2=5/8$, so $DE^2=5\sqrt{3}/2$.