Triangle Area Based on Segment Division

[nolachrymose]

Hi all,

I'm working on this problem in my Geometry class, and am having a bit of trouble coming to a conclusion. The problem as given is (I've attached the figure to this thread):

In this figure, <ACD is a right angle. A, B, and C are collinear, <A = 30, and <DBC = 45. If AB = 3 - sqrt(3), find the area of triangle BCD.

From this information, I have deduced that triangle DCB is right isosceles, with DC = CB. Also, since <DBC and <DBA are supplementary (by LPP), <DBA = 135, and thus, by Triangle Sum Theorem, <BDA = 15. Then, by AAP, this means that <CDA = 60, and thus triangle ADC is a 30-60-90 right triangle.

I've gotten to this point, and I know the ratios of this special triangle, but I'm having trouble forming the right equations. I was wondering if anyone could give me a lead onto which I could follow.

Thanks a lot! :)

[nolachrymose]

Nevermind, I figured it out! :)
For anyone who wants to know how I solved it, this is how I did it:

I assigned "x" as the length DC and CB. Since triangle ADC is a 30-60-90 right triangle, the hypotenuse (DA) must be 2x. Then, using Pythagorean theorem, I derived the following equation:

(2x)^2 - x^2 = (x + 3 - \sqrt{3})^2
4x^2 - x^2 = (x + 3 - \sqrt{3})^2
3x^2 = (x + 3 - \sqrt{3})^2
x\sqrt{3} = x + 3 - \sqrt{3}
x\sqrt{3} - x = 3 - \sqrt{3}
x(\sqrt{3} - 1) = 3 - \sqrt{3}
x = \frac{3 - \sqrt{3}}{\sqrt{3} - 1}
x = \sqrt{3}

From there, I used x^2/2 = area of triangle BCD, so I got 3/2.

By the way, how do I do multi-line LaTeX equations? I tried using the "\\" symbol, both at the end of the previous line and at the start of the next line, and neither worked? (I ended up using multiple [ itex ] tags to get the effect I wanted.)