- #1
nolachrymose
- 71
- 0
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):
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! :)
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! :)