Inscribed triangle in a semicircle

gmatblr wrote: We know DC = 25 / 4 .But we do not know BD (not sure how the solution guy assumes its value as 5)

How is the OA D?

Out of curiosity, who is the 'solution guy'? If the only information given is the information provided in the original post, the answer is certainly C and not D. If you take statement 2, for example, and AD=4, the diameter of the circle could be 5, or it could be 5,000,000- there's no restriction on how long DC could be. To see this, choose any circle with a large diameter AC, put the point D four units away from A on the diameter, and you have the picture given in the question. So CD could really be anything if you only know Statement 2. Of course the logic is the same for Statement 1. Clearly the two statements together are sufficient.

pepeprepa wrote:
We can write:
b/c=c/d=a/e
Which is also
bd=c^2
be=ca
And also:
c^2 + b^2=a^2
c^2 + d^2=e^2
a^2 + e^2=(b+d)^2

Finally, we have a,b,c,d,e which are unknowns and 5 inequalities, so we only need one value to find the others.
The thing is not to solve it but as soon as you have your ratios and your pythagore inequalities I think you have to choose D and that's all.

You can only solve five linear equations (and actually, those equations aren't linear, but that's not so important here) if the equations are independent. You have bd = c^2. Add the two equations:

c^2 + b^2=a^2
c^2 + d^2=e^2

b^2 + 2c^2 + d^2 = a^2 + e^2
b^2 + 2bd + d^2 = a^2 + e^2 (using the fact that bd = c^2)
(b+d)^2 = a^2 + e^2

So the equation (b+d)^2 = a^2 + e^2 is not a 'new' relationship among a, b, d and e, so doesn't count as an independent equation: we can derive this relationship from the other equations.