The following is presented within an exercise in H.E. Huntley’s The Divine Proportion. It’s not even the main point of the exercise; it’s only an assertion made in passing as the problem is presented, but I can’t figure it out.
On Page 42 of my copy, the author asserts the following:
We’re given a circle with a chord BA. We extend chord BA beyond the circle to a point P, such that, if we draw the tangent PT from P to a point T on the arc subtended by the chord BA, the tangential segment PT is equal in length to the chord BA. At this point the author asserts in passing that the following proportion exists among the various segments: PA/BA=BA/PB (or PT can be substitued for one or both BA’s, being equal to it).
Try as I might, I can’t see how this assertion is justified. I’m probably forgetting some basic lesson from high school geometry, which was a long time ago. Can anyone help?
PS…I’ve done my best to verbally describe a geometric diagram, but if you need clarification let me know and I’ll post it here.
I’m familiar with the book in question – I own a copy and have gone through it many times. I think I know what you’re asking.
The missing piece of the puzzle you need is that any angle you draw that subtends a given arc/chord must have the same angle. That is, if you start at one end of a chord and draw a line to any point on the circle’s circumference, them draw another line from the other end of the chord to the same point on the circumference, the angle between these two lines will always be the same, no matter which point on the circumference you choose (as long as it’s on the same side of the chord). The number of degrees in the angle is equal to one half the number of degrees in the arc.The classic case is that any triangle you draw inscribed within a circle that has a diameter as one side must be a right triangle. (It has to be a right triangle because the chord cuts right across the circle, so the chord subtends 180 degrees, half of which is ninety degrees).
In any event, use this tidbit along with the exterior angle theorem (if two straight lines cross the angles “diagonally” across from each other must be equal) and you see that the two triangles drawn inside the circle must be similar (not necessarily identical – you’ve shown that the angles are the same, but not that the sides are), so that corresponding sides must be in the same ratio. Voila! I’d be more explicit, but I’d have to draw a picture.
but I think it can be proved false quite easily. I will leave elegant demonstrations to others. I drew the figure and PT id not equal to AB. Or did I misunderstand the problem?
Calmeacham, I only follow you halfway. Of course, to substantiate the proportion given, I see that I have to show that Triangle(APT) is similar to (PBT), hence Angle(PTA) = Angle(PBT), but I don’t see how the law of equal inscribed angles subtended by a chord applies. Angle PTA doesn’t subtend the chord, does it?
javaman: Using your notation (and assuming I’ve understood the problem correctly) angle PTB and angle PAT are equal because they subtend the same arc of the circle. Angle TPA is shared and there you are.
This is known as the secant-tangent theorem. More generally, and changing notation, if a chord of length b is extended length a, and from that point a tangent is drawn of length c, then c/(a+b) = a/c.
>> We’re given a circle with a chord BA. We extend chord BA beyond the circle to a point P, such that, if we draw the tangent PT from P to a point T on the arc subtended by the chord BA, the tangential segment PT is equal in length to the chord BA
I have done this on a piece of paper and PT is NOT equal to AB. I just thought I’d let you know
I misread the problem. Thge point P is a given point which fulfills the condition that… Ok (boy is my face red) that’ll teach me to read more carefully…
Ignore the instruction to make PT equal to AB. In fact, ignore the tangency requirement.
Take a point outside the circle § and draw two lines that intersect the circle, one at points A and B, the other at points S and T. Then PA/PT = PS/PB, because of the similarity of the triangles PAT and PSB. (The two triangles share the angle at P, and the two angles at T and B are equal because they both subtend the same arc of the circle–therefore the two angles at S and A are equal, also.)
Now, just let S=T, and you have the original problem, too.
Thanks for your response. The original problem makes a lot more sense to me now. Unfortunately, I now have to demonstrate to myself the principle you cite, (and Calmeacham as well. It’s one of those things where if you draw the picture it looks right, but I need to discover for myself the logical proof.
To clarify (and keep someone from thinking you’re wrong like I did for a moment): you don’t “have the original problem” in the sense that this figure satisfies all the conditions (i.e. it doesn’t require that AB=PT), but you “have the oritinal problem” solved. A minor nitpick.
A nice solution. There is one thing that bugs me, about it, though: in the case where S and T are not coincident, you rely on the fact that angle STA is equal to angle SBA, because they both subtend arc SA. However, when S=T, you have a degenerate case where “angle STA” is undefined.
Look at it this way: Holding S and A still (that is, RM Mentock, on your figure which ignores the tangency requirement), examine the relationship between the position of T and the angle STA as T moves along the circle towards S. The function is constant until T crosses S, where there is a discontinuity. I’m not quite sure what this says about the relationship between the triangles STA and SBA at that point, but do notice that the moment T passes S, the triangles are no longer congruent.
I still believe it, but it’s not “straight from the book”, in Paul Erdos’ lingo.
That doesn’t work. The point of that step is “the two angles at T and B are equal because they both subtend the same arc of the circle”. If you substitute PTA when S=T, that step is no longer useful.
To demonstrate this, while S=T, move P. You have just changed angle PTA, but you haven’t changed angle SBA.
Nope. Remember that when S = T, the line PT is tangent to the circle at T. So you can only move P along the line which includes the original segment PT, and then only by simultaneously changing either A or B such that angle PTA = angle SBA. Given the points A, B, and S=T, point P is determined, so you can’t move it.
The two similar triangles are PTA and PBS. (In my picture point S is between P and T (unless S = T), and point A is between P and B, in case different pictures is causing confusion). Angle PTA = angle PBS holds even when S = T, and those angles are always well-defiend (assuming A, B, and T are distinct points).
A line going through S and T is only constrained to a particular direction if S and T are not coincident. When S = T, the line PT can point in any direction it wants, hence the discontinuity. I don’t see any other constraints making sure it remains tangent.
You are correct that as S and T come close together, the line going through them necessarily approaches being tangent to the circle because of the constraint, but once S = T, that constraint is gone.
I can easily pick a point P outside the circle, and points S and T which are coincident and on the circle, such that the line through all three points is not tangent to the circle.
But then that line will intersect the circle at some other point distinct from T (= S). This contradicts our definition of S and T. i.e. the point where the line intersects the circle which is closer to P is S, and the point where the same line intersects the circle which is farther from P is T. If S=T, then the line intersects the circle at only one point, but the only way for a line to intersect a circle at only one point is if the line is tangent to the circle at that point.
Ok, I’ll buy that. I was using a subtly different definition. Instead of S and T being “the two points at which the line intersects the circle”, I was considering them to be “two points which are both on the line and on the circle”. There is a difference.