I love problems like this because there are so many different approaches.
I must say though, I don't like Bunuel's approach, since he starts with the right answer, then shows that it is the right answer. But how did he know to start there in the first place?!
Let's work out the answer starting with what we know about the triangles.
Looking at \(\triangle\)ABC, we can express the area in two different ways:
\(Area = \frac{1}{2}AB*BC = \frac{1}{2}BD*AC\)
So \(AB*BC = BD*AC\) ...(1)
Since AC doesn't appear in the answer choices, we'll have to convert that into something else. Use Pythagoras:
\(AC^2 = AB^2 + BC^2\) ...(2)
Now let's combine equations (1) and (2)
First, square equation 1: \(AB^2*BC^2 = BD^2*AC^2\)
Then replace \(AC^2\) using equation (2): \(AB^2*BC^2 = BD^2*(AB^2 + BC^2)\)
Ok, now we have an expression involving only what we want, AB, BC and BD. We need to try to manipulate it into one of the answer choices. All the answer choices have BD on one side and AB and BC on the other, so we can start by doing that:
\(BD^2 = \frac{AB^2*BC^2}{(AB^2 + BC^2)}\)
This equation looks like a form we should be familiar with (from all those work rate problems). If we flip the equation over (take the reciprocal of both sides) it will look like:
\(\frac{1}{BD^2} = \frac{(AB^2 + BC^2)}{AB^2*BC^2}\) Now this is something we've seen many times. The equation can be rewritten as:
\(\frac{1}{BD^2} = \frac{AB^2}{AB^2*BC^2} + \frac{BC^2}{AB^2*BC^2}\)
\(\frac{1}{BD^2} = \frac{1}{BC^2}+ \frac{1}{AB^2}\) ...And this now matches one of our answer choices!
Answer B
_________________
Dave de Koos
GMAT aficionado