Notes — Chapter 13: The Concurrence Theorems

To the Student. “Once is chance, twice is coincidence, three times is enemy action,” Auric Goldfinger warns James Bond, as Bond crosses him again. If two lines in a configuration cross, that is not surprising. We expect it, but when three lines are concurrent at the same point, then we suspect something is going on. This chapter is about such surprises and why they happen. _ “ti

General Comments on Concurrences. When constructions lead to many lines all running through the same point, or many points that all fall on the same line, we should expect that there is some property that point has which makes all these lines run through it, or some special property of the line that forces all of our lines to run through it. This the pattern of most of the examples here.

There are other sorts of concurrence theorems. The Theorem of Pappus says that given rays r and s from point O and points A, B, and C on r and a, b, and con s, if you connect each of these points to the two points labeled by different letters on the other ray, the three intersections of these segments will be collinear. It is very easy to convince yourself by experiment that this is true just using paper and pencil. I did so in high school, but I was never able to discover a geometric proof of this fact.

In The Art of the Infinite, Oxford University Press, New York, 2003, Robert and Ellen Kaplan combine geometry and algebra to give a nice proof of the Theorem

of Pappus. Their proof suggests how to prove Pascal’s Theorem: If a hexagon is inscribed in a conic, then its opposite sides intersect in points that are all collinear. This is not exactly how they state Pascal’s result, but the results are equivalent. See their Chapter 8, Back of - Beyond.

These are theorems of projective geometry, so lines ordinarily considered to be parallel are regarded as meeting at points of an ideal line at infinity. There is a more geometric proof of the results of Pappus, Pascal, and Brianchon in Geometry and the Imagination by David Hilbert and Stephan Cohen-Vossen, in 2003 available as an American Mathematical Society Chelsea reprint, pages 104-106 and 117, and based on the geometry of hyperboloids, page 11. Pascal used algebra, not geometry, in his proof. The reason these results are true is because of the web of relationships that hold between the points and lines in the configuration. We should not expect some simple explanation, such as the fact that all of the perpendicular bisectors of chords of the same circle must be coincident because they all run through the center of the circle.

One projective result that has a direct geometric proof is Desargues’s Theorem. It says that if the lines through corresponding vertices of △ABC and AA’B’C’ are concurrent at some point O, then the intersections of the corresponding sides of these triangles are collinear on some line 2. See the exercise sequence Triangles in Perspective and its figures on page 622 of Geometry.

This is a projective theorem, so the point O or the line ℓ can be at infinity. To understand why this result must be true, we draw the diagram and think of realizing it as the projection into a plane of a pyramidal arrangement, with O as the apex of the pyramid. Then each of the triangles lies in a planar slice through this pyramid, and the line ℓ is the intersection of the planes making the slices that give triangles ABC and AA’B’C’. The intersections of the corresponding sides of these triangles lie in both these planes, and hence on @, as was to be shown.

In this chapter we begin with concurrences that have simple explanations and come to ones that require a more subtle approach, such as Ceva’s Theorem. Let the surprises begin!

Lesson 1. Triangles and Circles. The circle of standing stones at Stonehenge gets us going in this lesson. We see that every triangle has a unique circumcircle and hence is a cyclic polygon, one that can be inscribed in a circle. The perpendicular bisectors of the sides of a triangle are concurrent because they all run through the center of the triangle’s unique circumcircle.

li” Quit while You Are Ahead? Inductive Proofs. The ideas used to prove tat regular polygons are cyclic can be generalized to polygonal paths A Polygonal path with n segments, n > 3 is cyclic (all of its vertices lie

on a given circle O) if and only if the bisectors of any three of its segments (taken in order) are concurrent. The point of concurrence will be the center of the circle O. If a cyclic path includes all but one of the sides of a polygon, then the final side which joins the initial and final vertices of the path is also inscribed in O, so the polygon is cyclic.

This result is obvious from consideration of the result for triangles and its method of proof. Jacobs wisely kept to the case of triangles. If there were a situation where a student needed to make this generalization, the student could figure it out using the triangular case as a guide. An example often suffices to show us what is true, and why. It is clear that the method used for the triangle could deal with polygonal paths of 4 steps or more. Thus the general assertion above about polygonal paths being cyclic is clearly true by induction. But to give a formal proof by induction requires setting up the machinery. This induction argument is no more convincing than the example. We avoid the formalities, but some teachers may want to show their students induction at work in geometry. The above is one result that could be established by induction. A second is Theorem 74, page 573 of geometry: Every regular polygon is cyclic.

The Two Color Theorem. Here are two geometric results that can be proven by induction. Every map having a finite number of countries all bounded by doubly infinite lines (possibly intersecting) can be colored with two colors so that no two countries with a common border have the same color. The same is true if the boundaries are formed by complete (though possibly intersecting) circles. These results are easily proved by induction.

Induction is a powerful tool in geometry. The proof of the Four Color Theorem for planar maps was done by showing that any counterexample can be reduced to a counterexample with fewer countries (reducibility). This argument is equivalent to using complete induction, where you suppose the proposition is true for all k <n, and then prove to show that the proposition must be true for k = n — after showing that the initial cases are true, of course. Complete induction is equivalent to ordinary mathematical induction. “

lv Recursion/Induction and Geometry. Successive division of intervals into tenths, hundredths, and so on is an inductive or recursive process. Such cascade processes are essential for measurement — of lengths, areas, and volumes. Fractals are constructed by an inductive or recursive process: the Koch snowflake, the Sierpinski polygons (pages 144-145), flowsnakes, and dragon curves, to mention a few. You can find these things on the Web or in many books, including Benoit Mandelbrot’s book The Fractal Geometry of Nature, W. H. Freeman and Company, New York, 1982, where the idea of fractals was launched.

An interesting application of induction is working out tilings of the plane by rep-tiles, tiles that can be assembled in groups to form shapes similar to the basic tiles. See Reptiles exercises on page 422. As another example, four squares can be assembled to form a (similar) square at twice the original linear scale. Not very interesting, but let’s look farther. Take a square and mark it into four equal squares, then cut off the upper left-hand subsquare giving you an ¢-shaped tile. Four of these can be assembled into a larger similar ¢-shape, and so on, ad infinitum. The kite and dart tiles devised by Roger Penrose tile the plane, but do this in a way that has no translational symmetries (aperiodic tilings). Penrose tilings are usually created by using an inductive construction called inflation, in which the tiles are first enlarged and then the enlarged tiles are subdivided into tiles congruent to the original dart and kite tiles.

For rep-tiles, see Martin Gardner’s column of that title in Mathematical Games, SCIENTIFIC AMERICAN, May 1963, reprinted in The Unexpected Hanging by Martin Gardner, Simon and Schuster, 1969, available in 2003 from the University of Chicago Press. Also search the Web under rep-tiles, self-similarity, aperiodic tilings, and Penrose tilings. Other sources include Penrose Tiles to Trapdoor Ciphers by Martin Gardner, W. H. Freeman and Company, New York, 1989, which also covers Mandelbrot’s work and fractals. Tilings and Patterns by Branko Grtinbaum and Geoffrey Shephard, W. H. Freeman and Company, San Francisco, 1987, covers tilings in general, including aperiodic tilings. A shorter, paperbacked version was released later and does not cover aperiodic tilings.

Here is a tantalizing problem, unsolved in 2003. Is there a single tile that will tile the plane only aperiodically? This tile might be a rep- of frac- tile. This seems a simple question, but many have attacked it without success. Penrose found two tiles that only tile the plane aperiodically. Can we get it down to one, or is two the best that can be done? There is a lot on the Web about tilings, especially about beautiful or unusual tilings. Look at David Epstein’s Geometry Junkyard on the Web or search on names such as John Conway, Charles Radin, together with subjects such as tiling, aperiodic, rep-tile (or reptile), or fractile. t

Set I begins with a light review of the ideas. RGB Color, Exercises 15-17 feature construction, discovery, and proof, while Exercises 18-21 feature construction and discovery. Four Lines and Four Circles, Exercises 22-25 gives us an astonishing concurrence of four circles arising from a seemingly random configuration of four lines and their six points of intersection. This is a consequence of Miquel’s Theorem, which can be easily proven after we know the properties of cyclic quadrilaterals. A Web search for Miquel’s Theorem will bring a rich harvest. A wonderful source for Miquel’s Theorem and other concurrence results is Ross Honsberger’s Episodes

in Nineteenth and Twentieth Century Euclidean Geometry, New Mathematical Library from the Mathematical Association of America, 1995. Honsberger is an artist. He makes everything beautifully simple and he has an eye for “mathematical gems.”

Because there is so much about Miquel on the Web, I will be brief here. In the figure at bottom left, page 533 of Geometry, the key is △ADC. Points B, F, and E are chosen along the lines determined by the three sides of △ADC. By Miquel’s Theorem, the red, orange, and green circles, which are called Miquel circles for this system, are all concurrent. A Miquel circle for a triangle and the points chosen, one on each of the lines determined by the sides of the triangle, is a circle through one of the vertices of the triangle and through the two chosen points that lie on sides running through that vertex. I leave you with the mystery of why the circumcircle for △ADC must be concurrent with the three Miquel circles. This configuration is shown with no hints in the delightfully puzzling Penguin Dictionary of Curious and Interesting Geometry by David Wells, Penguin Books, 1991.

Nearest School (30-32) yields nice construction and discovery exercises. This is called Fermat’s problem, and the point that solves it is the Fermat point. In the book cited above, David Wells gives the following physical interpretation of the Fermat point. Take three equal weights tied to strings, one passing frictionlessly through a hole at each vertex of the triangle. Tie all three strings together in a knot and keep the triangle in a horizontal plane. The weights will pull the strings toward each vertex with equal force. The position of equilibrium for the knot will be the Fermat point, which leaves as little string as possible on the plane of the triangle and puts as much of the string length below the plane of the triangle as possible. This puts the center of mass of the weights as low as it can be. Three equal forces in a plane can only be in balance if they make 120° angles with each other. Students could build a model as a project to see how this works. This interpretation is also given in Hugo Steinhaus’s Mathematical Snapshots, Third Edition, Oxford University Press, 1983. This book, in any edition (try Dover Publications), has many wonderful applications, mostly presented visually with minimal text. It is highly recommended.

Set ITI. Folding Experiment (33-40) is a nice exercise in construction, discovery, and measurement. Babylonian Problem (46-48) includes construction, measurement, and comparison to exact results. Diameters and Sines (49-53) asks students to draw conclusions, give their reasons, and do calculations.

Set III. Slipping Ladders is an old but good problem. The answer is intuitive at one level but surprising at another. I doubted my initial insight until I checked it.

Lesson 2. Cyclic Quadrilaterals. A convex quadrilateral is cyclic if and only if it has a pair of Opposite angles that are supplementary. I have restated the theorem to emphasize the condition given in the text that the quadrilateral must be convex. The difficulty arises because our angles are ordinarily no larger than 180°, so if you take a cyclic quadrilateral and flip one of its vertices so that it lies an equal distance from its neighbors, but is inside the circle instead of being on it, that angle and its opposite one are still supplementary — according to our usual way of measuring things.

The proof of the converse part of this theorem on page 537 shows that the locus of points D such that ∠ADC is a constant (say k) is the arc of a circle. Actually, we get two arcs for this locus, one on each side of the line AC. This description of circular arcs is useful. You can use our results on secant angles to prove this. If D moves off the given arc, the angle at D, as calculated using intercepted arcs changes from the value, k, of the angle when D is on the arc.

Set I. These exercises are straightforward and interesting. A Different Proof (13-17) is so nice Jacobs kept it even though one reviewer felt that one proof was enough. In Equilateral Triangles on the Sides (24-28) the point P happens to be the Fermat point for △ABC. From arguments using angle and arc measure it is clear that ∠APC = 7CPB = ∠BPA = 120°. It is this condition that forces all of these circular arcs to go through P. This depends on the locus property for circular arcs and angles noted above.

Set II. Cyclic or Not? (29-31) combines visual thinking and reasoning and therefore, proof. Bramagupta’s Theorem (32-38) combines construction, discovery, and proof. Area Formula (39-41), also due to Bramagupta, would make a nice project for exploration using Geometer’s Sketchpad™. The library or on the Web might shed light on the history of this result. Ptolemy’s Theorem (42-55 and on to 57) and its corollaries were used by Ptolemy to work out his table of chords. Specializing it so that AB is a diameter and the quadrilateral is an isosceles trapezoid, you can get the angle addition formula for the sine. See John L. Heilbron’s Geometry Civilized, Oxford University Press, Oxford, 1998, pages 162 and 163.

Set III. There are two other such sets of points. See Jacobs’s answer section. This could lead to a nice class discussion. Why did some students find one set but not the other set of points?

Lesson 3. Incircles. Every triangle has an incircle, whose center is determined by the intersection of any two of its (consequently) concurrent angle bisectors. The results on incenters are exactly analogous to those on circumcenters in Lesson 1. These results generalize in the same way. If n > 3, a n-gon P has an incircle if and only if the bisectors of any three consecutive angles of P are concurrent.

If the center of the incircle for P is O, we start at some vertex A and find O as the intersection of the angle bisectors at neighboring angles A and B. Then our condition tells us that the angle bisector for the next point, C, around P also runs through O, and so on. If Z is the last vertex before we return to A, and all the other angle bisectors run through O, then the bisector of Z must run through O as well. The incircle we have from looking at earlier vertices is tangent to sides YZ and ∠A, so the ray ∠O bisects angle Z.

Set I. Exercises 1-9 cover the basic ideas. Properties of circles and tangents are used in Exercises 10- 19. Construction Problem (27-30) shows that two noncongruent triangles can have the same incircle and circumcircle. Poncelet’s porism says that if an n-gon can be simultaneously circumscribed about circle O and inscribed in circle Q, then no matter what point you start from on Q, you can simultaneously inscribe an n-gon in Q and have this same n-gon circumscribed about O. See Eric Weisstein’s MathWorld entry on this topic on the Web. In 2003 there was an animation of the porism on this site.

▶ Porisms. A related result that can be understood completely using circular inversions is Steiner’s porism. Here is Steiner’s construction: Given circle O inside circle Q, construct a chain of circles starting with Co, tangent to O and Q. Each C(;41) in the chain Co,Ci, Co,… is tangent to O, Q, and C;. Steiner’s porism is that this chain either closes up with some C,, just touching Co as well as O, Q, and C,,,_1) or it never closes up. Which of these alternatives happens depends on O and Q but is independent of the position of Co.

Steiner’s porism can be proved by using a circular inversion to reduce the initial situation to a pair of concentric circles, for this configuration the assertion of the porism is obvious. Since inversions preserve circles and lines, this special symmetric case allows you to show that the general version of the porism holds. This is a comment of H. S. M. Coxeter’s from his Introduction to Geometry, Second Edition, John Wiley and Sons, New York, 1969, page 87._ il

▶ Porisms and Technology. These porisms cry out for animations, which have been created using various software, generally Geometer’s Sketchpad™. Search on keywords such as Java applet, hexlet, bow] of integers, Steiner chains, seven circles theorem. Students will find many things that could be the basis for projects in this area.

Set II. Ship Location, Exercises 31-39. The data suggest the ship is located on each of these lines ℓ,, 2, and é3, but they are not concurrent. The point P, which is equally close to @;, @2, andé3 is a good candidate for our estimate of the ship’s true position. The results found in Incircle Problems 1 (40-47) and 2 (48-58) are explained by Incircle Problem 3 (59-62). Trisector Challenge, 63, is easy enough if you don’t get distracted by angle trisection.

Set III. A nice exercise. See Jacobs’s comments in his teacher materials.

Lesson 4. The Centroid of a Triangle. For finding balance points we can treat a solid body as if its mass were concentrated at one point, its center of mass. The center of mass of a uniform triangle is its centroid. The medians of a triangle are concurrent at the triangle’s centroid. Because a median bisects each infinitesimal strip of the triangle that crosses it parallel to the side that the median bisects, each of these strips will balance perfectly, if allowed to pivot independently about the median. So the triangle as a whole is perfectly balanced if allowed to pivot on a median.

Suppose we knew that a body has a center of mass about which it balances perfectly on any axis. Then we can conclude that that point must be at the intersection of any two medians of a uniform triangle. This is a physical argument for the concurrence of the medians of a triangle. See Medians as Bisectors, Exercises 38-46. The geometrical proof is found in Exercises 5-12.

It is somewhat surprising that the concurrence of the medians can be used to prove that the altitudes of a triangle are concurrent. The intersection of the altitudes is the orthocenter. See Exercises 13-19.

Set I. The proofs of Theorems 71 and 72 (5-19) are cleanly outlined and standard. Median Construction (20-26) involves construction and discovery strands. Altitude Construction (27-31) involves construction, visual thinking, and discovery.

Set II. Medians as Bisectors (38-46) tells us why triangles balance along their medians. Doing Without a Compass (47-54) is a beautiful application with an element of discovery. Engineering Challenge (55 and 56) is a nice challenge.

Set III. Balancing Point. The answer is within reach. Discovering why it is right, possibly using Geometer’s Sketchpad™ or other aids would be an interesting project.

▶ Centers of Mass, Medians, and Ceva’s Theorem. A more flexible way to think of this is to suppose that masses m,, Mp, and, mc are at the vertices of △ABC. If we are working with coordinates or thinking of these points as vectors, the center of mass (C.M.) of this arrangement will be

C.M. for △ABC = =-(maA +mpB+mcC),

where M = ma +mp+mc > 0. Archimedes did not have algebra and vectors to aid him. He used his Law of the Lever, which states that the center of mass of the weights at B and C will be at a point Y on segment BC such that mp/mc = CY/YB. This comes down to saying that the masses at B and C act as a single mass Mpc = ™Mp+mMe at the point Y, which turns out to be

the point we would write in vector or coordinate form as

Y =C.M. for masses at B and C

= assaf ar mcC).

MBC Then Archimedes would find the center of mass for all three masses by locating the center of mass for the system consisting of the mass ma at A and the mass mpc at Y, again using his Law of the Lever.

This calculation is

CM: for △ABC = (maA + mpcY).

Mma +™MBC

This tells us that the center of mass of a triangle △ABC with masses at its vertices lies on the segment from A to the Y, the center of mass of the masses on side BC opposite from A.

Take points X, Y, and Z on the sides of △ABC as follows: X is on side AB, Z is on side AC, and Y is on side BC, and choose these so that they satisfy the equation below, which is the hypothesis of Ceva’s after clearing the fractions. See the next lesson in Geometry, Ceva’s Theorem, page 555, top, for the formula and the figure.

(*) AX. BY -CZ =AZ-BK«CY,

Next define masses ma, mp, and mc by

ma =BX-CY, mp =AX- CY, and me = AX- BY.

Using (*) and Archimedes’ Law of the lever, algebra shows that

Y = center of mass of mg and mo,

X = center of mass of mg and ma, and

Z = center of mass of ma and mc.

So the Cevians are the lines from each vertex through the center of mass of the masses on the other vertices. Thus they are concurrent because they all run through the center of mass of the three masses we assigned to the vertices.

The concurrence of the medians and of the Cevians can be looked at purely geometrically, as Jacobs does in Geometry. Alternatively, these results can be looked at physically using centers of mass, or algebraically, as I have suggested, using vectors. Each point of view has advantages and drawbacks. If you have vector machinery at hand, the vector approach is simplest, because it Just uses calculation — which is easier than the inspiration geometric proofs require. Developing familiarity with vector methods would divert Jacobs from his main task. So he chose a geometrical approach.

Lesson 5. Ceva’s Theorem. This is an extension of the result that the medians of a triangle are concurrent. The note above shows how it is related to centers

of mass. The approach in the text is direct and geometric.

The facts of this lesson are more important than the proof, and more memorable. If students go on and use this result or learn other approaches to it, the geometric proof will be less important than knowing that the theorem is true and knowing that there are several ways to look at it.

Andrew M. Gleason noted that the Ceva Condition is conveniently expressed and used in this alternative form

AZ CYB Ka A Xe BYICZ:

Start at any vertex then take the product of every other segment going clockwise for one side of the equation, then do the same thing going counterclockwise from the starting vertex to get the other side of the equation. See Ollies Equation (25-27), for variants on the usual form of the Cevian condition.

In the forward direction of the proof of Ceva’s Theorem, you show that a (AX/BX)(BY/YC)(CZ/AZ) = 1 if the Cevians are concurrent. Given X and Z, we know by algebra that there is only one Y on the side BC such that

1 *) (BY/YC) = AX7BX)(CZ/AD)]

We know that we will find a Y’ that satisfies () by drawing the line BP, where P = CX BZ. So we know Y = Y’. This is the proof of the converse, and it is a corollary of the fact that the algebraic condition () determines the position of Y when the positions of the other points are given.

The converse of the Pythagorean Theorem has this same character. It seems that when you prove the converse, you are repeating what you have already established. You are, but with a twist. This can puzzle students, who think they have already proven the converse, and with good reason.

Set I. Which is Which? (1-5) covers the basics: recognizing things and naming them. Using Ceva’s Theorem (6-9) gives students practice in seeing and solving. A calculator will be useful here. Equilateral Triangle (10-−14) requires construction, measuring, and conjecture.

Set II. The Theorem of Menelaus (28-33) gets us off to a nice start. Concurrent or Not? (34-36) uses Ceva’s Theorem to establish a generalization of the theorem that the medians are concurrent. The result in What Kind of Triangle? (37-47) seems surprising. The Gergonne Point (48-51) is a nice place to use Gleason’s form of the Ceva Condition. Another Look at Altitudes (52-59) gives us another proof that the altitudes of a triangle are concurrent. It is also a nice review of similarity arguments.

Set III. Area Puzzle. This is remarkable, and the more so for there being two ways to arrive at the right answer.

Lesson 6. Napoleon’s Discovery and Other Surprises. There are an absolutely staggering number of results that involve concurrences, constructions of various figures on the sides of other figures, special points relating to a triangle, and so on. Ross Honsberger’s Episodes in Nineteenth and Twentieth Century Euclidean Geometry, New Mathematical Library from the Mathematical Association of America, 1995 is a wonderful source of these things. Also search on the Web for such topics as The Nine Point Circle, Symmedian Point, Euler Line, and Clark Kimberling’s Triangle Centers. Other sources are H. S. M. Coxeter’s Introduction to Geometry, Second Edition, John Wiley & Sons, New York, 1961, and Geometry Revisited by Coxeter and S. L. Greitzer, The Mathematical Association of America, 1967, and Robert and Ellen Kaplan’s The Art of the Infinite, Oxford University Press, New York, 2003.

The Euler line is covered in Exercises 6-34. This sequence is broken into parts that can be assigned separately. The lead-in for this lesson is covered in Exercises 61-63, Napoleon Triangles. Set III, Alternating Triangles appeared as the discovery exercise, Equilateral Triangle Mystery, on page 275. The rest of Sets I and II have their surprises and rewards.

Set III. The secret of proving that the natural conjecture here is true is to stmplify. The sides of the resulting parallelogram are parallel because they are parallel to one or the other of the diagonals of the original quadrilateral. So if you draw only the parts of the diagram needed to check this, and if you draw one of the diagonals, then things are simpler. Once simplified, you can see why one of the sides is parallel to one diagonal. Then you have the solution.

▶ Circles and Triangles. The diagram, bottom left, on page 565 of Geometry shows the situation for Napoleon’s Theorem. An alternative approach can be based on observing that the circumcircles of the equilateral triangles are concurrent. This observation allows you to dispense with segments EB, DA, and CF. The fewer lines the better. The circles through CBD and CAE intersect at a point P. The inscribed ∠CPB and ∠CPA are both 120°, so it follows that ∠APB must also be 120°, and hence the circle through ABF also passes through P. This is because the locus of all points X such that X lies on the same side of the line AB as point C and ∠AXB = 120° is the arc of a circle. The locus of points on the opposite side of line AB from C and such that ∠AYB = 60° is the other arc of this same circle, because △XBY is a convex quadrilateral and so ∠AXB + ∠AYB = 180°, as we expect. So E is on the other arc of this circle as shown.

Now we need to show ∠ZYX = LYXZ = LKZY = 60° so that △XYZ is equilateral. The argument is to

observe that XY bisects both the minor arcs from A towP; sovZAYG = ∠PYG? Similarly, ∠BYH= ∠PYH. So ∠AYG+ ∠BYH = ∠PYG + ∠PYH, because equals added to equals give equals. The sum of these four angles is the central angle ∠AYB. Because ∠AYB is the central angle for the inscribed angle ∠AF’B (= 60°), we know, AY Bi==120%. )So.2ZY X=: ∠DYiEz=1ZP YG ∠PYH = 60°. And the same argument shows that the other angles of △XYZ are 60° as well, so △XYZ is equilateral. tl

Summary and Review. There is a mix of review and discovery in these exercises. All can be done at the level asked, but exercises like Five Circles (28-31) may challenge some students to go beyond what is asked for and find proofs of some amazing patterns. All is fair in going after such problems. After all, Euler used algebra to find his line. The geometric proof came later. Explorations with Geometer’s Sketchpad™., library research, or Web browsing may be needed to unravel some of these mysteries. The example below is a case in point.

ll” Going at It Hammer and Tongs. The Five Circles exercises would be a baffler were it not for the comments Jacobs in the draft materials for the teacher. Looking at the diagram on page 568, we have the cyclic quadrilateral ABCD in a circle we will call O. Let X be the midpoint of AB, Y the midpoint_of BC, Z the midpoint of CD, and W the midpoint of DA. Let Q be the intersection of the chords XZ and YW of O.

/XQY = 5(B + BY + 2D + BW) =

7(B + 6F + 70+ + D+ HW+ WA+ AX) =

360° = 90°, because AX = XB, BY =Yc, etc.

So XZ and YW are perpendicular. Let S be the center of the incircle of △ABC and T be the center of the incircle for △ABD. We show that reflection in line XZ interchanges S and T. We locate S by taking the intersection of the bisectors of ∠ACB and ∠BAC. If you draw the diagram with A, X, B, Y, C, Z, D, and W, and mark the arcs that are equal, you will see that these bisectors are AY and XC.

We locate T, the incenter of △ABD in a similar way as the intersection of angle bisectors BW and XD. Reflection the line XZ interchanges segments AY and BW and it interchanges segments XC and XD, hence it interchanges points S and T. By the same sort of argument, reflection in line XY interchanges the incenter of ABCD with that of △ACD.

Applying like reasoning to reflection in the line YW we find that these four incenters are permuted by reflection in this line as well. Hence the set of these four incenters is invariant under reflections in the two perpendicular lines XZ and YW. Hence these four points are the vertices of a rectangle that has XZ and YW as lines of symmetry.