FOR CHAPTER 2, PAGE 47: TO PROVE THAT AN ANGLE BISECTOR DIVIDES THE OPPOSITE SIDE PROPORTIONALLY TO THE TWO ADJACENT SIDES.

We begin with triangle ABC with angle bisector AT_a (see figure A-1, figure 2-5, and figure 5-3). We construct BP parallel to AT_a, meeting CA extended at point P. From the parallel lines, we have ∠CAT_a = ∠APB, and ∠T_aAB = ∠ABP. However, . Therefore, ∠APB = ∠ABP, and triangle ABP is isosceles; thus AB = AP. From the parallel lines we have . Replacing AB for AP, we get the sought after result: .

 

FOR CHAPTER 3, PAGE 69:
NAPOLEON'S THEOREM:
IF EQUILATERAL TRIANGLES ARE CONSTRUCTED ON THE SIDES OF ANY TRIANGLE (EITHER OUTWARD OR INWARD) THE CENTERS OF THOSE EQUILATERAL TRIANGLES THEMSELVES FORM AN EQUILATERAL TRIANGLE.

To prove this theorem we consider ΔACB′ (see figure A-2 and figure 3-5). Since Q is the centroid (point of intersection of the medians) of ΔACB′, AQ is two-thirds of the length of the altitude (or median). Using the relationships in a 30°-60°-90° triangle, we find that .

Similarly, in equilateral ΔABC′,

Therefore, AC:AQ = AC′:AR. Also ∠QAC = ∠RAC′ = 30°, ∠CAR = ∠CAR (reflexive) and therefore, by addition, ∠QAR = ∠CAC′.

We can then conclude that ΔQAR ~ ΔCAC′. It follows that CC′:QR = .

Similarly, we may prove , and .

Therefore, BB′:PQ = AA′:PR = CC′:QR.

But since BB′ = AA′ = CC′ (as proved earlier), we obtain PQ = PR = QR.

Thus we can conclude that ΔPQR is equilateral.

FOR CHAPTER 5, PAGE 125: DERIVATION OF STEWART'S THEOREM.

This theorem yields a relation between the lengths of the sides of the triangle and the length of a cevian of the triangle (see figure 5-19): a(d² + mn) = b²m + c²n.

In ΔABC, let BC = a, AC = b, AB = c, CD = d. Point D divides BC into two segments; BD = m and CA = n. Draw altitude AE = h and let DE = p.

In order to proceed with the proof of Stewart's theorem, we first derive two necessary formulas. The first one is applicable to triangle ABD.

We apply the Pythagorean theorem to triangle ABE to obtain AB² = AE² + BE².

However, by applying the Pythagorean theorem to triangle ADE, we have AD² = AE² + DE², or d² = h² + p², also h² = d² – p².

Replacing h² in equation (I), we obtain

A similar argument is applicable to triangle ACD.

Applying the Pythagorean theorem to triangle ACE, we find that AC² = AE² + CE².

However, h² = d² – p², so we substitute for h² in (III) as follows

Equations (II) and (IV) give us the formulas we need.

Now multiply equation (II) by n to get

and multiply equation (IV) by m to get

Adding (V) and (VI), we have

b²m + c²n = d²m + d²n + m²n + mn² + 2mnp – 2mnp,

therefore b²m + c²n = d²(m + n) + mn(m + n).

Since m + n = a, we have b²m + c²n = d²a + mna = a(d² + mn), which is the relationship we set out to develop.

FOR CHAPTER 5, PAGE 131:
PROOF FOR THE SUM OF THE DISTANCES FROM ANY POINT IN A TRIANGLE IN TERMS OF THE SIDE LENGTHS.

AP² + BP² + CP² = AG² + BG² + CG² + 3GP²

 

Begin by letting Q be the midpoint of AG (see figure A-3). We now apply the earlier-developed relationship about the medians to each of the following triangles (see chapter 5, p. 129).

Substituting in equation (III) and multiplying by 2 we get

 

Now adding (I), (II), and (IV):

 

A similar argument made for median BM_b yields

 

For median CM_c we get

 

By adding (V), (VI), and (VII):

 

We now apply an earlier-developed relationship to triangle ABC (see chapter 5, p. 130), that

 

3.(AG² + BG² + CG²)= AB² + AC² + BC².

 

Now substitute this into equation (VIII) to get our desired result:

 

AP² + BP² + CP² = AG² + BG² + CG² + 3GP².

 

FOR CHAPTER 5, PAGE 133: THE CENTROID AS BALANCING POINT.

AX = BY + CZ

Draw medians AM_a, BM_b, and CM_c (see figure A-4). From Q, the midpoint of AG, draw PQ YZ. Also draw RM_a YZ. Since ∠AGX = ∠RGM_a, and AQ = GQ = GM_a (property of a centroid). ΔGRM_a ΔGPQ, also RM_a = PQ. AX||CZ, therefore, RM_a is the median of trapezoid BYZC, and (property of median of a trapezoid). PQ = AX (property of a midline). Therefore, (transitivity), and AX = BY + CZ.

FOR CHAPTER 6, PAGE 136: PROOF OF MENELAUS'S Theorem AZ ⋅ BX ⋅ CY = AY ⋅ BZ ⋅ CX, IF AND ONLY IF, X, Y, AND Z ARE COLLINEAR.

To prove that if X, Y, and Z are collinear, then AZ ⋅ BX ⋅ CY = AY ⋅ BZ ⋅ CX.

Draw a line containing C, parallel to AB, and intersecting XYZ or YXZ at D (see figures A-5a and A-5b). We are thus beginning with the given collinear points X, Y, and Z.

From equations (I) and (II), , from which we easily get AZ ⋅ BX ⋅ CY = AY ⋅ BZ ⋅ CX.

Now we shall prove that if the points X, Y, and Z are so situated (with one of the points on the extension of the sides of the triangle) that the equation AZ ⋅ BX ⋅ CY = AY ⋅ BZ ⋅ CX is true (or another way of expressing this is that , then the three points X, Y, and Z are collinear.

We will let the intersection point of AB and XY be the point Z′. Then we have to prove Z′ = Z.

Because of part 1 (above) we have also . Therefore is Z′ = Z, and the points X, Y, and Z must be collinear.

FOR CHAPTER 6, PAGE 137: PROOF OF SIMSON'S THEOREM (USING MENALAUS'S THEOREM) THE FEET OF THE PERPENDICULARS DRAWN FROM ANY POINT ON THE CIRCUMSCRIBED CIRCLE OF A TRIANGLE TO THE SIDES OF THE TRIANGLE ARE COLLINEAR.

We begin by drawing PA, PB, and PC (see figure A-6).

, . Therefore ∠PBA = ∠PCA = α.

Thus, = cot (in ΔBPZ and ΔCPY), or ,

 

Similarly, .

Therefore, (in ΔAPZ and ΔCPX), or ,

Since ∠PBC and ∠PAC are opposite angles of an inscribed (cyclic) quadrilateral, they are supplementary. However, ∠PAY is also supplementary to ∠PAC.

Therefore ∠PBC = ∠PAY = γ.

Thus, (in ΔBPX and ΔAPY), or ,

 

By multiplying (I), (II), and (III) we obtain

 

Thus by Menelaus's theorem, X, Y, and Z are collinear. These three points determine the Simson line of triangle ABC with respect to point P.

 

FOR CHAPTER 6, PAGE 152: THE CENTER OF THE NINE-POINT CIRCLE OF A TRIANGLE IS THE MIDPOINT OF THE SEGMENT FROM THE ORTHOCENTER TO THE CENTER OF THE CIRCUMCIRCLE.

Besides the orthocenter H, the centroid G, and the circumcenter O, we have the nine points on the nine-point circle with center N (see figure A-7):

M_a, M_b, M_c — midpoints of the sides of triangle ABC,

H_a, H_b, H_c — feet of the altitudes of triangle ABC,

E_a, E_b, E_c — Euler points of triangle ABC (that are the midpoints of the segments between the orthocenter H and the vertices of triangle ABC).

Since E_cM_c subtends a right angle at point H_c, it must be the diameter of the nine-point circle. Therefore, the midpoint, N, of E_cM_c is the center of the nine-point circle.

Extend AO through O to intersect circumcircle O at point D. Then draw BD and CD. OM_c is a midline of triangle ABD. Therefore, OM_c||BD. Since∠ABD is inscribed in a semicircle, it is a right angle. Now both BD and CH_c are perpendicular to AB, so that BD||CH_c. Similarly, CD||BH_b.

We therefore have parallelogram CDBH, so that BD = CH. Also, OM_c = BD (OM_c is a midline of triangle ABD).

Therefore, OM_c = CH = E_cH, and OM_cHE_c is a parallelogram (one pair of sides both congruent and parallel). Since the diagonals of a parallelogram bisect each other, the midpoint, N, of E_cM_c is also the midpoint of OH.

THE LENGTH OF THE RADIUS OF THE NINE-POINT CIRCLE OF A TRIANGLE IS ONE-HALF THE LENGTH OF THE RADIUS OF THE CIRCUMCIRCLE.

In figure A-7, we notice that E_cN is a midline of triangle OHC. Therefore, E_cN = OC, which justifies the above statement.

In a paper published in 1765, Leonhard Euler proved that the centroid, G, of a triangle trisects the segment OH, that is, This line, OH, is the Euler line of a triangle. (See figures 6-8 and 6-9.)

THE CENTROID OF A TRIANGLE TRISECTS THE SEGMENT FROM THE ORTHOCENTER TO THE CIRCUMCENTER.

We have already proved that OM_c||CH (see figure A-7), and we have proved that OM_c = CH.

We then have ΔOGM_c ~ ΔHGC (AA) with a ratio of similitude of . Therefore OG = GH, which may be stated as .

It now remains for us to prove that G is the centroid of triangle ABC. From the triangles we just proved similar, we have

But since CM_c is a median, G must be the centroid, because it appropriately trisects the median.

[An alternative proof is very short and elegant, when we use the dilation D(G, – ) with center G and dilation factor –. Here triangle M_aM_bM_c is the image of triangle ABC. Therefore the altitudes of triangle M_aM_bM_c are the images of the altitudes of triangle ABC. That means the image of the orthocenter H is the intersection point O of the perpendiculars; also H, G, and O are collinear, and we have HG = 2 GO.]

It is interesting to note that

We can then see that HG is divided internally by N and externally by O in the same ratio. This is known as a harmonic division.

ALL TRIANGLES INSCRIBED IN A GIVEN CIRCLE AND HAVING A COMMON ORTHOCENTER ALSO HAVE THE SAME NINE-POINT CIRCLE.

Because all triangles inscribed in a given circle and having a common orthocenter also must have the same Euler line, the center of the nine-point circle for all these triangles is fixed at the midpoint of the segment OH on the Euler line (established above). Since the radius of the nine-point circle for each of these triangles is half the length of the circumradius (see above), they all have their nine-point circle with the same radius as well as a fixed center. Thus they all must have the same nine-point circle.

TANGENTS TO THE NINE-POINT CIRCLE OF A TRIANGLE AT THE MIDPOINTS OF THE SIDES OF THE TRIANGLE ARE PARALLEL TO THE SIDES OF THE ORTHIC TRIANGLE. (See p. 154.)

Radius NM_c of the nine-point circle is perpendicular to tangent DM_c (see figure A-8). The circumradii, which contain the vertices of the triangle ABC, are perpendicular to the corresponding sides of the orthic triangle H_aH_bH_c. Therefore, OC H_aH_b.

We showed earlier that E_cN is a midline of triangle OHC, and therefore, E_cN||OC.

This implies that E_cNM_c||OC. Thus DM_c||H_aH_b. The proof for the remaining two sides of the orthic triangles is done in the same manner as that shown above.

TANGENTS TO THE NINE-POINT CIRCLE AT THE MIDPOINTS OF THE SIDES OF THE GIVEN TRIANGLE ARE PARALLEL TO THE TANGENTS TO THE CIRCUMCIRCLE AT THE OPPOSITE VERTICES OF THE GIVEN TRIANGLE.

Since the tangents to the circumcircle at a vertex of the triangle and the tangents to the nine-point circle at the midpoints of the sides of the triangle are each parallel to the sides of the orthic triangle, they are parallel to each other.

An orthocentric system consists of four points, each of which is the orthocenter of the triangle formed by the remaining three. In figure A-8, the points A, B, C, and H form an orthocentric system, since

 

H   is the orthocenter of triangle ABC,

A   is the orthocenter of triangle BCH,

B   is the orthocenter of triangle ACH, and

C   is the orthocenter of triangle ABH.

 

(See p. 154.)

THE FOUR TRIANGLES OF AN ORTHOCENTRIC SYSTEM HAVE THE SAME NINE-POINT CIRCLE.

The proof of this property is left to the reader, since all that is required is to check to see if, for each of the four triangles, the nine determining points all lie on the same circle N (see figure A-8, and see figures 6-20 and 6-21).

THE NINE-POINT CIRCLE OF A TRIANGLE IS TANGENT TO THE INCIRCLE AND THE EXCIRCLES OF THE TRIANGLE.

N is the center of the nine-point circle, I the center of the inscribed circle, H the orthocenter, and O the center of the circumscribed circle (see figure A-9, and see figures 6-11 and 6-22; see p. 156):

 

F_i — point of tangency with the inscribed circle (at I),

F_a — point of tangency with the escribed circle (at side a),

F_b — point of tangency with the escribed circle (at side b),

F_c — point of tangency with the escribed circle (at side c).

 

The proof of this property is quite complex and time consuming. The interested reader will find four different proofs of Feuerbach's theorem in Modern Geometry, by Roger A. Johnson ([Boston, MA: Houghton Mifflin, 1929], pp. 200–205).

The proof that Feuerbach actually used consists of computing the distances between the center of the nine-point circle and the centers of the inscribed circle (r), the circumscribed circle (R), and the circle inscribed in the orthic (or pedal) triangle, H_aH_bH_c (radius p), and showing that they equal the sum and difference of the corresponding radii.

 

OI² = R² – 2Rr (Euler)

IH² = 2R² – 2Rp

OH² = R² –4Rp

 

[N.B. I is the center of the inscribed circle, G is the centroid, H is the orthocenter, and O is the center of the circumscribed circle.]

 

FOR CHAPTER 6, PAGE 158: PROOFS OF MORLEY'S THEOREM: IN ANY TRIANGLE, THE THREE POINTS OF INTERSECTION OF THE ADJACENT ANGLE TRISECTORS FORM AN EQUILATERAL TRIANGLE. (See chapter 6, figure 6-24.)

We present three proofs.

PROOF 1:

(Y. Hashimoto, “A Short Proof of Morley's Theorem,” Elemente der mathematik 62 [2007]: 121.)

Hashimoto's original proof is short and without a figure:

Let α, β, and γ be arbitrary positive angles with α + β + γ = 60°. For any angle η we put η' = η + 60°. Let triangle DEF be an equilateral triangle, and A [resp. B, C] be the point lying opposite to D [resp. E, F] with respect to EF [resp. FD, DE] and satisfying ∠AFE = β', ∠AEF = γ' [resp. ∠BDF = γ', ∠BFD = α'; ∠CED = α', ∠CDE = β']. Then ∠EAF = 180° – (β'+ γ') = α, and similarly ∠FBD = β, ∠DCE = γ. By symmetry it is enough to show that ∠BAF = α and ∠ABF = β as well.

The perpendiculars from F to AE and BD have the same length s. If the perpendicular from F to AB has length h < s, then ∠BAF < α and ∠ABF < β. If, on the other hand, h > s, then ∠BAF > α and ∠ABF > β. Since ∠BAF + ∠ABF = α' + β'+ 60° – 180° = α + β, we see that necessarily h = s and ∠BAF = α, ∠ABF = β.

PROOF 2:

Hashimoto's proof with a figure:

Let α, β, and γ be random angle measures, with α + β + γ = 60°.

For each of these angles we set the following: α' ≔ α + 60°, β' ≔ β + 60°, γ' ≔ γ + 60°. We have the triangle P′Q′R′ as equilateral. On each of the sides of the triangle P′Q′R′ we draw the triangles Q′A′R′, R′B′P′, and P′C′Q′ so that the following angles are produced:

 

The perpendicular from R′ to A′Q′ and B′P′ have the same length, s. If the perpendicular distance from R′ to A′B′ is h, where h < s, then the following is true: φ = ∠R′A′B′ < α and Ψ = ∠R′B′A′ < β.

On the other hand if h > s then it follows that

 

PROOF 3:

(L. Bankoff, “A Simple Proof of the Morley Theorem,” Mathematics Magazine 35, no. 4 [1962]: 223–24.)

From here, and similarly for triangles BXZ and CXY. It thus follows that the sum of angles around X, excluding ∠YXZ is 300°, or ∠YXZ = 60°. The other two angles are similarly shown to be 60°.

FOR CHAPTER 7, PAGE 168: [BEFORE FIGURE 7-9] DERIVATION OF HERON'S FORMULA FOR THE AREA OF A TRIANGLE.

There is nice geometric proof—actually attributed to Heron—provided in Thomas Heath's A Manual of Greek Mathematics (New York: Dover, 1963). However, to conserve space we will provide a much shorter trigonometric derivation building on the two high-school-level relationships: The law of cosines, c² = a² + b² – 2ab cos C; and the Pythagorean identity, sin²θ + cos²θ = 1. We had just previously developed the area of a triangle formula as Area ΔABC = ab sin C.

With cos² 1and substituting for sin C in this last equation for the area, we get

We now have to factor the term under the radical sign:

 

FOR CHAPTER 9, PAGE 278: ERDÖS-MORDELL INEQUALITY

PA + PB + PC ≥ 2(PD + PE + PF)

Our proof is based on one by H. Lee (“Another Proof of the Erdös-Mordell Theorem,” Forum Geometricorum 1 [2001]: 7–8), and it applies Ptolemy's theorem (see figure 9-19 and p. 278).

We begin the proof by drawing HB and GC perpendicular to line HFEG. BC is either equal to HG (if BHGC is a rectangle) or is greater than HG, if not. Symbolically, BC ≥ HG = HF + FE + EG.

Since AFPE is a cyclic quadrilateral,1 ∠APE + ∠AFE (angles at the circumference). However, ∠AFE = ∠BFH. Therefore, ∠APE = ∠BFH, thereby determining that ΔBFH ~ ΔAPE.

This gives us

Using this procedure, we can show that

The famous Ptolemy theorem,² which for quadrilateral AFPE gives us

PA ⋅ FE = AF ⋅ PE + AE ⋅ PF, can be rewritten as

With appropriate substitutions, we get

 

This can be rewritten as

 

PA ⋅ BC ≥ PE ⋅ BF + AF ⋅ PE + AE ⋅ PF + PF ⋅ CE

= PE (BF + AF) + PF (AE + CE) = PE ⋅ AB + PF ⋅ AC.

Dividing by BC, we have

 

If we were to use the other two sides of triangle ABC from which to draw the perpendiculars as we did for BC, we would get and .

Using the fact that for positive real numbers m and n it follows that , we can conclude that PA + PB + PC ≥ 2(PD + PE + PF). Of course, if and only if triangle ABC is equilateral and P is the center of the circumscribed circle then PA + PB + PC = 2(PD + PE + PF).

FOR CHAPTER 9, PAGE 282: TO PROVE THAT

We begin with right triangle ABC, where∠C = 90°, ∠A = α , and ∠B = β (see figure A-3). Also recall that AB = c, BC = a, and AC = b.

For the proof we will use the trigonometric functions cos and cos It then follows that

 

The equality only holds true when a = b, which would be an isosceles right triangle. Considering that c < a + b, we get from the above inequality that