How Do Strangely Aligned Capacitor Plates Affect Distance Calculations?

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mateomy
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Working along step-by-step with a text and I'm stumped as to how they reached a conclusion. In the first picture you see two plates aligned at an unspecified angle, theta. One plate is at a potential V and the other is grounded. The distance between the plates at one end is d, and at the other end d+a. You can see in the second picture how they've set it up with respect to a two dimensional coordinate system. Maybe it's late and I'm braindead or maybe not in any even I can't see how they've formed the relationship between b,d and a to determine the distance between the actual origin and the intersection of the plates (dotted lines in picture three). Can anybody enlighten me?

Thanks.

Perspective diagram of two parallel rectangular plates, the upper one tilted, with separation d at one end and d+a at the other, width w and length b labeled, and x, y, z axes shown.


Diagram of two plates meeting at apex O' on the negative x-axis, the upper plate rising from height d+a at x = b to the y-axis, with point (x, y), distance r, and angles theta and theta-0 marked.


Two equations: the length OO' equals bd over a, and theta-zero equals arctan of a over b.
 

Discussion

P.S. the whole point is to find the potential, but this is where I'm hung up.
 
In the triangle (in x<0) with base OO' and the height d, you have
tan(theta_0)=d/(OO').

In the trapezoid (in x>0), draw the horizontal at height d to form a similar triangle,
with base b and height a. Thus
tan(theta_0)=a/b.