The lines and points of a triangle combined with the circles that either enclose them or are enclosed by them are the key to many secrets embedded in triangles. In this chapter we will provide some of the most fascinating and surprising relationships of these triangle-related parts. Many of these relationships were not known to Thales (ca. 624–ca. 546 BCE), Pythagoras (ca. 570–ca. 510 BCE), Euclid (ca. 300 BCE), and Archimedes (ca. 287–ca. 212 BCE), our forefathers in this visual part of mathematics. As a matter of fact, Euclid in his famous book, Elements, mentions the centers of the inscribed and circumscribed circles of a triangle and with them the concurrent lines that determine these center points, namely, the angle bisectors of the three angles of the triangle and the perpendicular bisectors of the sides of the triangle, respectively. It was not until Archimedes's contributions that there was mention of the altitudes and the medians of a triangle. Till about the end of the eighteenth century, only five significant points relating to the triangle were known. However, not until the nineteenth century did geometry begin to blossom through the work of some famous mathematicians including Joseph-Diaz Gergonne (1771–1859), Jakob Steiner (1796–1863), Karl Wilhelm Feuerbach (1800–1834), Christian Heinrich von Nagel (1803–1882), Joseph Jean Baptiste Neuberg (1840–1926), and many others. More on these mathematicians later! Meanwhile, today there are more than 3,600 noteworthy points related to the triangle. There are many more discoveries about triangle parts that constantly appear in professional journals under “new discoveries.” We hope that the reader's exposure to this chapter will entice further exploration and perhaps even some new discoveries.

INTRODUCTION TO THE ALTITUDES OF A TRIANGLE

When one thinks of lines related to a triangle—other than its sides—one generally thinks of the altitudes,1 angle bisectors,² and medians.³

Aside from the properties that define these special line segments, each of these groups of three are concurrent line segments and each point of concurrency is a significant point in the triangle.

In figure 2-1, we have the three altitudes of triangle ABC. Beside the fact that they are perpendicular to the sides of the triangle, we can take note that the point H of concurrency—called the orthocenter of the triangle— partitions each of the altitudes into two line segments, the product of which is the same for each of the three altitudes. That is, for triangle ABC in figure 2-1: AH ⋅ DH = CH ⋅ FH = BH ⋅ EH. This evolves from the similarity of triangles in this configuration. That is, since

 

ΔAFH ~ ΔCDH, . This gives us AH ⋅ DH = CH ⋅ FH.

Similarly,

ΔAEH ~ ΔBDH, therefore, , and then AH ⋅ DH = BH ⋅ EH.

In figure 2-2, where we renamed the feet of the altitudes with subscripts, we can express this relationship as

AH ⋅ HH_a = BH ⋅ HH_b = CH ⋅ HH_c.

This leads to a rather unusual relationship in the triangle. If we form rectangles from the two parts of each of the cut altitudes, we find that they determine three equal-area rectangles, as shown in figure 2-3.

In figure 2-4, we once again show a triangle with altitudes drawn, and their “feet” (that is, the point at which the altitude intersects with the base) marked as H_a, H_b, and H_c, and the segments along the sides as a₁ = BH_a, a₂ = H_aC, b₁ = CH_b, b₂ = H_bA, c₁ = AH_c, and c₂ = H_cB. This allows us to state easily a relationship discovered in 1828 by the Swiss mathematician Jakob Steiner (1796–1863),⁴ namely, that a₁² + b₁² + c₁² = a₂² + b₂² + c₂².

INTRODUCTION TO THE ANGLE BISECTORS OF A TRIANGLE

The angle bisectors of a triangle, besides bisecting the angles of the triangle, meet at a point that is equidistant from each of the sides of the triangle, and is, therefore, the center of the circle inscribed in the triangle as shown in figure 2-5. In other words, the (perpendicular) distance from this point of intersection—known as the center of the inscribed circle or incenter—to the three sides of the triangle is the same for all three sides: IP_a = IP_b = IP_c, where P_a, P_b, and P_c are the feet of the perpendiculars from the incenter and the points T_a, T_b, and T_c are the points of intersection of the angle bisectors and the opposite side. Another way of looking at this configuration is to note that the circle is tangent to each of the three sides of the triangle. Later, we will justify that the incenter, I, of this circle is in fact the point of intersection of the angle bisectors. We should also note that AT_b = AT_c, BT_c = BT_a, CT_b = CT_a.

Every triangle has an inscribed circle (inside the triangle) and three escribed circles (outside the circle). Each of these circles is tangent to the three sides of the triangle. These are also shown in figure 2-6 for a randomly drawn triangle. These four circles are sometimes called equicircles. The centers I_a, I_b, and I_c of the escribed circles are determined by the intersection of two external-angle bisectors and one internal-angle bisector.

Once again, we have a neat relationship determined by the feet of the perpendiculars (P_a, P_b, and P_c) from the incenter to the sides of the triangle. Using the designations shown in figure 2-7, namely that

a₁ = BP_a , a₂ = P_aC , b₁ = CP_b , b₂ = P_bA , c₁ = AP_c , c₂ = P_cB,

we get a₁² + b₁² + c₁² = a₂² + b₂² + c₂².

INTRODUCTION TO THE MEDIANS OF A TRIANGLE

The medians of a triangle, which join a vertex of a triangle with the midpoint of the opposite side, trisect each other at their point of concurrency. That is, in figure 2-8 the following is true:

AG = 2 ⋅ GM_a
BG = 2 ⋅ GM_b
CG = 2 ⋅ GM_c

Furthermore, the point G is the center of gravity, or the centroid, of the triangle—that is, the balancing point. If you have a cardboard triangle, and you want to balance it on the point of a pencil, then the point at which it would balance is the centroid.

We can prove the mutual trisections of the three medians through the geometry taught in high school. We will show one possible method here. However, the power of knowing some geometric theorems beyond those taught in high school will come in very handy, as we shall show a bit further on. We will show a very simple procedure using elementary methods.

We begin by drawing parallel lines to CM, and going through points A, M_b, M_a, B, as shown in figure 2-9.

For triangle ACM_c we have AM_b = CM_b; therefore, AU = UM_c. In a similar fashion, we can eventually show that AU = UM_c = , and BV = VM_c = .

We have AU = UM_c = M_cV, and it follows that AD = DG = GM_a = . This relationship can be simply repeated for each of the other two medians, which in turn would allow us to establish concurrency at the trisection point—the centroid.

INTRODUCTION TO THE PERPENDICULAR BISECTORS OF A TRIANGLE

The perpendicular bisector p_c of the line segment AB is a line that is perpendicular to AB and contains its midpoint M_c (see figure 2-10). Every point, P, along the perpendicular bisector of AB is equidistant from the endpoints of AB, so that AP = BP. The three perpendicular bisectors of a triangle ABC, p_a, p_b, and p_c, are concurrent at the point O, which is the center of the circumscribed circle of triangle ABC.

This can be easily justified in that the point O is the intersection of p_a and p_b (see figure 2-11). They form the property of equidistance from the endpoints, so we get (for p_a) BO = CO and (for p_b) AO = CO. Therefore, AO = BO. Consequently, point O must also be on the perpendicular bisector p_c. Thus, the three perpendicular bisectors are concurrent at point O.

INTRODUCTION TO CEVA'S THEOREM: A SIMPLER WAY OF PROVING CONCURRENCY

When the altitudes, angle bisectors, and medians are presented in a high-school geometry course, the proofs of their concurrency that are provided are not at all simple. These proofs are often bypassed for convenience. Yet, if one delves into some geometry beyond what is presented in high school, there is a theorem that makes proving these three concurrencies much simpler. This very powerful and useful theorem was first published by Giovanni Ceva (1647–1734) in 1678 in his De lineis invicem secantibus statica construction. Ceva's theorem is an equivalence—or biconditional—which means that the converse is also true. To justify it requires two proofs—the original statement and its converse. For us to accept this theorem it would be appropriate to prove its validity. Ceva's theorem states the following:

The three lines containing the vertices of triangle ABC (figure 2-12) and intersecting the opposite sides in points L, M, and N, respectively, are concurrent if and only if .

There are many proofs available to justify this theorem, yet we shall use just one of these methods to prove this wonderful theorem. It is perhaps easier to follow the proof looking at the left-side diagram and then verifying the validity of each of the statements in the right-side diagram. In any case, the statements made in the proof hold for both diagrams.

To prove the “only if” part of the theorem, we consider figure 2-12, where we have on the left triangle ABC with a line (SR) containing A and parallel to BC, intersecting CP extended at S and BP extended at R.

The parallel lines enable us to establish the following pairs of similar triangles:

Now by multiplying (I), (II), and (V) we obtain our desired result:

This can also be written as AM ⋅ BN ⋅ CL = AN ⋅ BL ⋅ CM. A nice way to read this theorem is that the product of the alternate segments along the sides of the triangle, made by the concurrent line segments, or cevians, emanating from the triangle's vertices and ending at the opposite side, are equal. (A line segment joining the vertex of a triangle with a point on the opposite side is called a cevian.)

Yet it is the converse of this statement that is of particular value to use here. That is, if the products of the alternate segments along the sides of the triangle are equal, then the cevians determining these points must be concurrent.

We shall now prove that, if the lines containing the vertices of triangle ABC intersect the opposite sides in points L, M, and N, respectively, so that , then these lines AL, BM, and CN are concurrent.

Suppose BM and AL intersect at P. Draw PC and call its intersection with AB, point N′. Now that AL, BM, and CN′ are concurrent, we can use the part of Ceva's theorem proved earlier to state the following:

But our hypothesis stated that

Therefore, , so that N and N′ must coincide, and thereby proves the concurrency.

For convenience, we can restate this relationship as:

 

If AM ⋅ BN ⋅ CL = MC ⋅ NA ⋅ BL, then the three lines are concurrent.

 

There is an interesting variation to the famous Ceva theorem, one discovered by the French mathematician Lazare Carnot (1753–1823). Here the concurrency of the three cevians is specified by the partitioned angles at the triangles vertices. In figure 2-13, we have cevians AL, MB, and CN that partition the angles as angle A into α₁ and α₂, angle B into β₁ and β₂, and angle C into γ₁ and γ₂.

They will meet at a common point P if and only if⁵ . The proof entails using the law of sines several times and can be found in various geometry books.⁶

USING CEVA'S THEOREM TO PROVE SOME FAMILIAR CONCURRENCIES

As we mentioned earlier, Ceva's theorem is not typically a part of the high-school course, yet it provides us with a very powerful—yet simple— tool for proving concurrencies. Follow along as we now prove the basic concurrencies that we encountered earlier, yet with very short proofs— using Ceva's theorem.

CONCURRENCY OF THE ALTITUDES OF A TRIANGLE

We can use Ceva's theorem to prove that the altitudes of triangle ABC (see figure 2-2) are concurrent.

In triangle ABC, where AH_a, BH_b, and CH_c are altitudes:⁷

Multiplying (I), (II), and (III) gives us

or AH_c ⋅ BH_a ⋅ CH_b = H_bA ⋅ H_cB ⋅ H_aC.

 

By Ceva's theorem, this indicates that the altitudes are concurrent.

CONCURRENCY OF THE ANGLE BISECTORS OF A TRIANGLE

To prove that the angle bisectors are concurrent using Ceva's theorem, we must rely on a relationship about angle bisectors introduced in the high-school geometry course, but also proved in the appendix, namely, that the angle bisector of a triangle divides the side to which it is drawn proportional to the two adjacent sides.

In figure 2-5, for angle bisector AT_a, we get .

For angle bisector BT_b, we get .

For angle bisector CT_c, we get .

Now multiplying these three equations, we get

which (by Ceva's theorem) allows us to conclude that the angle bisectors are concurrent.

CONCURRENCY OF THE MEDIANS OF A TRIANGLE

The proof that the medians are concurrent using Ceva's theorem is the simplest of all. We can see from the fact that each side is divided into two equal segments that the products of the alternate segments will be equal. Therefore, the medians must be concurrent. That can be seen in figure 2-14 (left side), where AM_c = M_cB, BM_a = M_aC, and CM_b = M_bA.

Since AM_c ⋅ BM_a ⋅ CM_b = M_cB ⋅ M_aC ⋅ M_bA, we can conclude (by Ceva's theorem) that the three medians are concurrent.

EXPLORING OTHER CONCURRENCIES

We are just beginning to scratch the surface of the almost-boundless concurrencies that can be found in a triangle. To begin our exploration, let's consider the medians of triangle ABC, as shown in figure 2-14 (left side), and notice that M_cM_b is parallel to BC. Suppose we slide M_cM_b toward point A, but keeping it parallel to BC, where P and Q are now the points of intersection with AB and AC, respectively, as shown in figure 2-14 (right side). We can show—using Ceva's theorem—that the three lines, BQ, CP, and AM_a are also concurrent (at X).

We already have PQ||BC.

By multiplying the equations marked (I) and (II), we get

Thus by Ceva's theorem, AM_a, QB, and PC are concurrent.

 

As we continue to explore concurrencies in a triangle, we shall consider the following very simple, yet surprising, relationship. The lines joining each of the vertices of a triangle and the point of tangency of the inscribed circle with the opposite side are concurrent. (See figure 2-15.) This very simple relationship was first published by the French mathematician Joseph-Diaz Gergonne (1771–1859). Gergonne reserved a distinct place in the history of mathematics as the initiator in 1810 of the first purely mathematical journal, Annales des mathématiques pures et appliqués. The journal appeared monthly until 1832 and was known as Annales del Gergonne. During the time of its publication, Gergonne had published about two hundred papers, mostly on geometry. Gergonne's Annales played an important role in the establishment of projective and algebraic geometry, as it gave some of the greatest minds of the times an opportunity to share information. We shall consider this rather simple theorem established by Gergonne, as it exhibits concurrency and is easily proved using Ceva's theorem in the following way.

In figure 2-15, circle I is tangent to sides BC, AC, and AB at points P_a, P_b, and P_c, respectively. It follows that AP_b = AP_c, BP_a = BP_c, and CP_a = CP_b. Each of these equalities may be written as

By multiplying these three fractions we get

Having now established that , by applying Ceva's theorem, we can conclude that AP_a, BP_b, and CP_c are concurrent at G. This point of concurrency, G, is called the Gergonne point of triangle ABC.

Suppose we now draw the triangle formed by joining the points of tangency of the inscribed circle. This triangle, ΔP_aP_bP_c, is often called a Gergonne triangle. Naturally, as for all triangles, the Gergonne triangle has many “centers.” One of these centers, or points of concurrency, is special in that it ties the Gergonne triangle to the original triangle. If we locate the three midpoints, M_a, M_b, and M_c, of the sides of the original triangle ABC, and from these points we draw perpendicular lines, DM_a, EM_b, and FM_c, to the sides of the Gergonne triangle, surprisingly, we find that these three perpendiculars are concurrent at the point P, as may be seen in figure 2-16.

We can now even take this a step further and find some more concurrency points in this geometric configuration. As in figure 2-17, instead of the midpoints (M_a, M_b, M_c) of the sides of the original triangle, we locate the midpoints D, E, and F of the circumscribed circle's arcs , and , respectively. Then we draw the lines joining these arc-midpoints with the respective tangency points P_a, P_b, and P_c of the inscribed circle with the triangle—which are the vertices of the Gergonne triangle. Surprisingly, we find these three lines (DP_a, EP_b, and FP_c) are concurrent at point Q. As an extra feature, we also notice that this point of concurrency (Q) is collinear (lies on the same line) with the center (I) of the inscribed circle and the center (O) of the circumscribed circle of the original triangle. The surprises keep on coming!

As if this were not enough, we also notice that there is another concurrency point, which we will find in the following manner. We begin by drawing the diameters of the inscribed circle from each of the three points, P_a, P_b, and P_c, of tangency (i.e., perpendicular to the sides) to meet the inscribed circle (whose center is the intersection of the perpendiculars at points P_a, P_b, and P_c) at points D, E, and F. Then, by drawing the lines joining each of these three points with the nearest vertex point of the original triangle (as shown in figure 2-18), we once again get a concurrency of lines. This time lines AE, BD, and CF meet at point S.

When we drew these diameters, there was some built-in symmetry and so the result was somewhat plausible. However, if we take this another step further and draw any three concurrent lines emanating from the tangency points (P_a, P_b, and P_c) with the inscribed circle, we amazingly get a similar concurrency just as before. That is, in figure 2-19, when DP_b, EP_a, and FP_c are concurrent at R, we have AE, BD, and CF concurrent at point T (see figure 2-19).

To further show how concurrency seems to emerge in unexpected places, we consider triangle ABC with the three concurrent (at point P) cevians AD, BE, and CF, from which we then draw triangle DEF and its inscribed circle. The points of tangency are, D′, E′, and F′, as shown in figure 2-20. Curiously enough, when we join these latter points (D′, E′, and F′) to each of their nearest vertices of triangle ABC, we once again have three lines that are concurrent—namely, AD′, BE′, and CF′ are concurrent at point P′.

SOME CONCYCLIC POINTS

It is well known that three noncollinear points determine a unique circle. When more than three points lie on the same circle we have a set of points that we tend to cherish, and we call them concyclic points. That is what happens when we draw lines through the Gergonne point, G, of a triangle and parallel to the sides of the Gergonne triangle as shown in figure 2-21, where DE || P_cP_b, FJ || P_cP_a, and HK || P_bP_a.

Quite unexpectedly, where these parallel lines intersect the sides of the triangle ABC is where we have the six points that lie on the same circle. Moreover, much more unexpectedly, that circle is concentric (shares the same center) with the inscribed circle of the triangle. This remarkable occurrence was first discovered by the German mathematician Carl Adams (1811–1849) in 18438 and, therefore, in his honor carries his name as the Adams' circle.⁹

MORE CONCURRENCIES

Recall that we arrived at the Gergonne triangle by using the tangency points of the inscribed circle of a triangle. In figure 2-22, we have the three escribed circles of triangle ABC. Each of these circles is tangent to the three sides of the triangle, yet each of these circles lies outside the triangle. Once again, as with the Gergonne point, the lines joining the tangency points with the opposite vertex of the triangle are concurrent at point N. So you can see there is a natural analog between the inscribed circle of a triangle and the escribed circles of the same triangle—in this case tied together by the Gergonne-point property.

This is not different from the point discovered by the German mathematician Christian Heinrich von Nagel (1803–1882). We can locate this Nagel point, N, of triangle ABC, in the following way:

Point P is situated on BC so that AB + BP = AC + CP (see figure 2-23).

Point Q is situated on AC so that BC + CQ = AB + AQ (see figure 2-24).

Point R is situated on AB so that BC + BR = AC + AR (see figure 2-25).

We then conclude that AP, BQ, and CR are concurrent. This point of concurrency is known as the Nagel point, N, of triangle ABC. We should note that this is analogous to the way we located the point determined by the tangency points of the escribed circles in figure 2-22.

While we are considering the escribed circles of a triangle, we should also appreciate that they give us another point, M, of concurrency, one that is often called the middle point (or mittenpunkt) of a triangle. It is determined by the three lines that join the centers of the escribed circles with the midpoint of the side of the triangle to which they are tangent.

In figure 2-26, we have points M_a, M_b, and M_c as the midpoints of the sides of triangle ABC. The lines PM_a, QM_b, and RM_c are concurrent at point M, which is the middle point of the triangle ABC. This middle point, M, is not identical with the Nagel point, N, of the triangle ABC.

CONCURRENCIES DETERMINING OTHER CONCURRENCIES

We now embark on another path toward exposing more secrets embedded in triangles. We will begin by drawing any three cevians that are concurrent and we will use this point of concurrency to find another concurrency point. This not only is unexpected but also shows how intertwined some characteristics of triangles can be.

We begin with triangle ABC, where we draw the cevian lines AD, BF, and CE so that they are concurrent at point P, as shown in figure 2-27. We then consider the circle drawn through the points D, E, and F, which also intersects the triangle in points D′, E′, and F′. Using Ceva's theorem, it can be shown that the lines AD′, BF′, and CE′ are also concurrent (at Q).

A somewhat analogous situation can be created as shown in figure 2-28, where we have triangle ABC with concurrent cevians AL, BM, and CN meeting at point P. Rather than use a circle through the feet of these cevians to determine the key points that we used above to determine another point of concurrency, we will construct parallels. We will draw NR||AC, LS||AB, and MT||BC to determine points R, S, and T. These newly determined points will determine yet another set of concurrent cevians, namely, AR, BS, and CT, which meet at point Q.

A similar case can be made for selecting parallel lines from the feet of the original concurrent cevians (AL, BM, and CN) to the other sides, namely, from point L parallel to AC instead of to AB, as we did earlier, from point M parallel to AB instead of BC, and from point N parallel to BC instead of AC. Actually, following this scheme, we can get lots more such points of concurrency of cevians by using each new set of concurrent cevians to generate another set of concurrent cevians. The reader may wish to explore these concurrencies further.

A similar type of reasoning will produce yet another rather unexpected concurrency. Suppose we have a circle intersecting each of the three sides of a randomly selected triangle twice, however, in such a way that the perpendiculars to the sides of the triangle at three of the points (D, E, and F) of intersection with the circle are concurrent (at P). (See figure 2-29.) When this is the case, we find that the perpendiculars to the three sides of the triangle at the remaining circle intersection points (K, L, and M) are also concurrent (at Q).

As if that weren't surprising enough, notice what happens when we consider the midpoint of the line segment joining these two points of concurrency. This is, in fact, the center, M, of the circle containing the six feet of perpendiculars.

We will now consider a somewhat more involved configuration—one that will consider midpoints of the sides of a randomly selected triangle, and the midpoints of a set of concurrent cevians of the triangle. These will curiously generate another set of concurrent cevians.

In triangle ABC (figure 2-30), the cevian lines AL, BM, and CN are concurrent at point P. Bear in mind that these cevians are any concurrent cevians of triangle ABC. We now locate the midpoints of AL, BM, and CN and label them as points D, E, and F, respectively. We also have points M_a, M_b, and M_c as the midpoints of the sides of triangle ABC. Using Ceva's theorem, it can be shown that, quite unexpectedly, DM_a, EM_b, and FM_c are also concurrent—at the point Q.

Having one set of concurrent cevians generating another set of concurrent cevians is always a striking curiosity. Consider the triangle ABC where the cevians AL, BM, and CN are concurrent at point P as shown in figure 2-31. The points D, E, and F are the midpoints of the line segments LM, MN, and NL, respectively. Surprisingly (and with the help of Ceva's theorem), the lines AD, BE, and CF can be shown to be concurrent at point Q. The triangle DEF will help the ambitious reader to justify this remarkable relationship.

To generate another unexpected concurrency we can extend the previous scheme. Again, we shall begin with triangle ABC, where the cevians AL, BM, and CN and concurrent at point P. Now, as shown in figure 2-32, suppose that the points D, E, and F are not (necessarily) the midpoints of NM, LN, and LM, respectively, but rather just three points on the triangle that determine concurrent cevian lines, LD, ME, and NF at point R. Astonishingly, these three points, D, E, and F, also yield similar concurrent lines as we had in the previous configuration. Namely, the lines AD, BE, and CF can be shown to also be concurrent at point Q, using Ceva's theorem.

Concurrency in a triangle seems to be endless. There are many more such concurrencies to be discovered, and we will encounter many more in the following chapters. However, the ambitious reader may want to seek other concurrencies at this point before reading on.