The following presentation contains graphical material.

As always, viewer Discretion is Advised.

Thm: Circles are to one another as the squares on their diameters.

Housekeeping:

Let ABCD and EFGH be circles, and let BD and FH be their diameters.

W.l.o.g. Assume cEFGH < cABCD

I will use c for circle, seg for segment of a circle, p for parallelogram,

and t for triangle.

Previous Propositions mentioned in the book:

IV.6

SAS congruence for triangles

III.17

From a given point to draw a straight line touching a given circle.

X.1

Given 2 unequal magnitudes, if from the greater, magnitudes greater

than half of the remainders are being removed, eventually, the

remainder will be less than the smaller magnitude.

XII.1

Similar polygons inscribed in circles are to one another as the squares

on their diameters.

V.11

Ratios which are the same with the same rario are also the same with one another.

V.16

If four magnitudes are proportional, then they are also proportional

alternately.

Archimedes Lemma (A.L.):

If A and B are positive magnitudes, then the numbers in the sequence

B, B/2, B/4,...will eventually be smaller than A.

Lemma

If S > cEFGH and S: cABCD = cEFGH:T (*), then cABCD > T.

Ok,since we need to use this lemma, might as well justify it now:

By V.16, we can rewrite the equation (*)as follows:

S:cEFGH = cABCD:T, and since S > cEFGH, T < cABCD follows.

Now, let's prove the thm.

Pf by contradiction:

Suppose cEFGH:cABCD ne FH^2:BD^2

So, we have two cases:

Case I: cEFGH:cABCD > FH^2:BD^2

Case II:cEFGH:cABCD < FH^2:BD^2

from these two cases, we can have more cases.

Case I: S:cABCD = FH^2:BD^2, for S < cEFGH or S > cABCD

Case II: S:cABCD = FH^2:BD^2, for S > cEFGH or S < cABCD

Lets first consider

S:cABCD = FH^2:BD^2, for S < cEFGH.

look at the pictures in the postscript file.

first show that sEFGH = 1/2 big square

look at the pictures again.

Since we know sEFGH < cEFGH < big square,

and big square = 2*sEFGH,

cEFGH < 2*sEFGH, so sEFGH > 1/2 cEFGH.

Next, show tENH = 1/2 p

look at the pictures one more time.

Since we know tENH < seg ENH < p and tENH = 1/2 p,

segENH < 2 tENH, so 1/2 segENH < tENH.

(A.L and X.1) By bisecting the remaining circumferences of cEFGH and

joining straight lines repeatedly to make polygons, we will eventually

have some segments of the circle < cEFGH - S.

Thus, we will get a polygon >S! I call it poly1.

Now, keeping this fact in mind, we can inscribe a polygon similar to

poly1 in cABCD; I named it poly2. By XII.1, we have the following ratio:

poly2:poly1 = BD^2:FH^2

from before, we have cABCD:S = BD^2:FH^2

By V.11, we have poly2:poly1 = cABCD:S

By V.16, we have cABCD:poly2 = S:poly1

Since cABCD > poly2, S > poly1, but we have shown earlier the opposite.><.

For the S < cABCD case, just replace cABCD with cEFGH.

The S > cEFGH case is quite short.

We have cABCD:S = BD^2:FH^2, which is the same as s:cABCD = FH^2:BD^2.

By lemma, S:cABCD = cEFGH:T, for some T < cABCD, but this is

the same as the previous case.

For the S > cABCD case, just replace cABCD with cEFGH in the previous case.

The post script file which contains 22 images related to the theorem can be downloaded by clicking the file name: project.ps (32 kb) |

O.K. I am done!

Questions?

Thanks to Professor Casselman and Professor Austin for providing the topic and helping me throughout the process of this project.

Thanks to Weidong, Wenjun, Myung-Sin, Peter, Gerhard, Christine, Maria,Professor Fix, and David for technical support.

Thanks to Adam for proof-reading.

Thanks to http://aleph0.clarku.edu/~djoyce/java/elements/bookXII/propXII2.html for the proof.