I’ve been reading an interesting book by H.E. Huntley, The Divine Proportion. (Interestingly, although the book was published in 1970, the entire content seems to be available on Google Books.
For some time I’ve been looking at a construction on page 42. (The red line segments have been added by yours truly). [ul]
[li] The chord BA is drawn on the circle. [/li][li] Line PT is constructed to be the same length as the chord BA, and so as to be tangent to the circle at T.[/li][li] Chord BA is extended to meet line PT at P.[/li][/ul]
According to the book, the above conditions mean that:
AP/BA = BA/PB (or, by definition, AP/PT=PT/PB).
Since
BA=AP - BA
this can be rewritten as
BA^2=AP(AP-BA)
and after a little algebraic fiddling,
(AP/AB)^2 - (AP/AB) - 1 = 0,
Which tells us that (AP/AB)=Φ (by the definition of Φ).
Now I do get the algebraic part of it, but I’m having trouble grasping the geometry. How do we know that AP/BA=BA/PB? If my knowledge of geometry hasn’t entirely dissipated, for that to work, triangle APT has to be similar to triangle PBT. Obviously, both triangles share the vertex and angle at P, so that’s one angle down and one to go (because if I prove that two angles are similar, then the third follows automatically). But I can’t figure out how to prove that the two triangles share a second angle. An issue of major concern is that I would expect the tangentiality of PT to be a necessary component of the derivation, but I can’t see how that would be.
This is all correct. Here is a relevant fact: The inscribed angle (sometimes also called an interior angle) of a circle is half the included arc. In your diagram, angles PTB and PAT are inscribed angles of the same arc, TB.
Evidently there’s something I still need to learn about what “inscribed angle” means, but no doubt your link will help. PAT (or BAT) is obviously inscribed in the arc, but I don’t see how that’s true of the other angle.
Oh, I see your problem now. The angle formed between a chord and a tangent (like BT and PT) is a degenerate case, which is only mentioned in the Wikipedia article but not really explained. If you imagine that PT is not-quite-tangent to the circle, but instead intersects it at two points T[sub]1[/sub] and T[sub]2[/sub], very close together (with T[sub]1[/sub] closer to P), then you can see that angle T[sub]1[/sub]T[sub]2[/sub]B=PT[sub]2[/sub]B is an inscribed angle on arc T[sub]1[/sub]B. You can probably convince yourself that things behave nicely in the tangent limit.
(Alternately, of course you can prove directly that the statement holds when one of the lines is a tangent, but the above argument makes it easy to see why the cases ought to be the same.)
That wasn’t meant to be a proof; that was just motivation to draw a connection between the two cases.
In intro (high-school-level) geometry, the statement (an inscribed angle is half the included arc) would probably be stated and proved as two separate theorems, one where the angle is formed between two secant chords and one with the angle between a secant and a tangent, with no epsilons harmed in either proof (geometry is traditionally taught before calculus, around here at least).
