Alfred S. Posamentier - The Secrets of Triangles: A Mathematical Journey

The authors wish to extend sincere thanks for proofreading and useful suggestions to Dr. Michael Engber, professor emeritus at the City College of the City University of New York; Dr. Manfred Kronfeller, professor of mathematics at Vienna University of Technology, Austria; Dr. Bernd Thaller, professor of mathematics at Karl Franzens UniversityGraz, Austria; and Dr. Peter Schpf, professor of mathematics at Karl Franzens UniversityGraz, Austria. We are especially grateful to Dr. Robert A. Chaffer, professor of mathematics at Central Michigan University for contributing on fractals. We also thank Catherine Roberts-Abel for managing the production of this book and Jade Zora Ballard for her meticulous editing.

FOR : TO PROVE THAT AN ANGLE BISECTOR DIVIDES THE OPPOSITE SIDE PROPORTIONALLY TO THE TWO ADJACENT SIDES.

We begin with triangle ABC with angle bisector ATa (see and TaAB = 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 :
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 ACB (see ). Since Q is the centroid (point of intersection of the medians) of ACB, 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 ABC,

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

We can then conclude that QAR ~ CAC. 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 PQR is equilateral.

FOR : 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 ): a(d2 + mn) = b2m + c2n.

In ABC, 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 = d2p2.

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 = d2p2, 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 :
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 ).

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