Alternative Geometric Proofs of Theorems for Con cyclic Points for a Triangle

(1)

Mem. Fae. Educ., Kagawa Univ. II, 56(2006), 51-59

Alternative Geometric Proofs of Theorems for Con cyclic Points for a Triangle

by

Hiroo FUKAISHI, Kazunori FUJITA and Akiko MATSUSHIMA

(Received July 14, 2006)

Abstract

In this paper we present alternative geometric proofs of the theorems for concyclic points for a triangle given in [5].

§

1. Introduction

In [3 - 5] we have developed a drawing game on a display as a teaching material for elementary geometry classes with activities using computers. In this paper we present alternative geometric proofs of the theorems for concyclic points for a triangle given in [5].

For terminology of geometry throughout the paper consult [1, 2, 6, 7].

As for the present paper, the first author gave the geometric proofs of the theorems. The last author wrote the programs for all the figures to the theorems under the direction of the second author.

(2)

§ 2. Geometric Proofs of Theorems

Theorem 1. ( [5; Theorem 1]) In MBC let S be the center of the circle determined by the incenter I and the excenters lb, le in the interiors of LB, LC, respectively. Then we have the following :

(1) (The radius of the circle S)

=

2 x (the radius of the circumcircle of MBC), (2) The excenter Ia in the interior of LA, the circumcenter O and the point

S are collinear, (3) Ia O

=

OS (Fig. 1).

Proof Take the points A1, B1, C1 on the extensions of the line-segments IaA, IaB, Ia ^C, respectively, such that

IaA=AA1, IaB=BB1, IaC=CC1.

Then AABC and AA1B1C1 are in a position of similarity with the center Ia and the similitude ratio 1 : 2. The circumcenter O of MBC corresponds to the circumcenter 0, of M1B1Ci by the similarity.

To complete the proof of Theorem 1 it suffices to show that the circumcenter 01 of M1B1Ci coincides with the center S of the circle determined by the points I, lb, le.

i) The circle 01 passes through the point /.

In fact,

LA1IC1 = LIIaC1 + LIC1Ia

= 2LllaCi by AIIaC= AIC1C,

= 2 LIBC since the quadrangle /Bia C is inscribed,

= LABC

= LA1B1C1 by MBC ^Cl)AA1B1Ci.

Since both points B1, I are in the same side of the line A1C1, the circle 01 passes through the point I by the converse of the Theorem of inscribed angles.

(3)

Alternative Geometric Proofs of Theorems for Concyclic Points for a Triangle

ii) The circle 01 passes through the point lb.

In fact,

LAJbCJ

=

^LJbA1A⁺^LJblaA

=

²^LhlaA

=

²^LIIaC

=

^LA1B1C1 by the proof of i).

Since both points Bi, lb are m the same side of the line A1C1, the circle 01 passes through the point lb by the converse of the Theorem of inscribed angles.

iii) The circle 01 passes through the point le.

Similar proof is omitted.

D

Theorem 2. ( [5; Theorem 2]) Let S be the center of the circle determined by the vertices B, C and the orthocenter H of LABC.

If

the circle S passes through the circumcenter O and the incenter I, then we have the following :

(l) HI= IO,

(2) The circle S passes through the point la, (3) Hla = Ia 0,

where Ia denotes the excenter in the interior of LA (Fig. 2).

The assumption in Theorem 2 that the circle S passes through both points 0 and / is superfluous; but it is posed for the convenience of algebraic calculations by a computer. We have the following :

Lemma 1. Let S be the center of the circle determined by the vertices B, C and the orthocenter H of LABC.

If

the circle S passes through the circumcenter 0, then it does through the incenter L

Proof Note that LA

<

^{90°. Let} ^{D, E, F} be the feet of the perpendiculars from the vertices A, B, C to the opposite side, respectively.

Since the quadrangle AFHE is inscribed in a circle, LBHC = 180° - LA.

By the Theorem of inscribed angles,

(4)

LBOC= LBHC.

2LA

=

180° - LA.

:. LA= 60°.

Hence LBIC = LBOC = 120°.

Since. both points 1, 0 are in the same side of the line BC, the circle S passes through the point / by the converse of the Theorem of inscribed angles.

D

Lemma 2. Let S be the center of the circle determined by the vertices B, C and the orthocenter H of i6.ABC.

If

the circle S passes through the incenter I, then it does through the circumcenter 0.

Proof By the Theorem of inscribed angles,

LBHC

=

^LBIC, ^or ^LA

=

^60°.

Hence LBHC = LBIC

=

120°.

Since . both points /, 0 are in the same side of the line BC, the circle S passes through the point O by the converse of the Theorem of inscribed angles.

D

Proof of Theorem 2. We have LA

=

60° by the proof of Lemma 1.

i) i6.0BS is an equilateral triangle.

In fact, we have i6.0BS

=

i6.0CB, since OB = OC and SB = SC.

Because LBOC

=

^120°,

·. we have LBOS

=

60° and OS = BS = OB.

· Thus, the circumcircle O of i6.ABC passes through the point S.

ii) The quadrangle OBSC is a rhombus.

· Hence the diagonals OS and BC bisect each other and intersect at the right angles.

iii) The quadrangle AHSO is a rhombus.

Let A', B' be the midpoints of the sides BC and CA of MBC, respectively. Then i6.HAB and i6. OA 'B' are in a position of the similarity with the center G and the similitude ratio 2 : 1.

Since AH II OS and AH = OS = OA, the quadrangle AHSO is a rhombus.

iv) LISH

=

LISO.

Since the diagonal AS of the rhombus AHSO bisects LHAO and LBAH

=

L CAO, the line AS bisects LBAC so that it passes through the

· incenter /. This implies iv) .

(5)

( 1) Both arcs fii and

W

on the circle S are congruent by iv) ; that is, HI

=

^IO.

To see (2) and (3) consider the point Ia instead of the point I.

D

Theorem 3. ( [5; Theorem 3]) Let S be the center of the circle determined by the vertices B, C and the circumcenter O of .6.ABC.

If

the circle S passes through the incenter I, then we have the following :

( 1) The circle S passes through the orthocenter H, (2) HI = IO,

(3) Hia

=

Ia 0, where Ia denotes the excenter in the interior of LA (Fig. 3) . Theorem 4. ( [5; Theorem 4]) Let S be the center of the circle determined by the vertices B, C and the orthocenter H of MBC.

If

the circle S passes through the incenter I, then we have the following :

( 1) The circle S passes through the circumcenter 0, (2) HI

=

^IO,

(3) Hla = Ia 0, where 1a denotes the excenter in the interior of LA (Fig. 4).

Theorem 5. ( [5; Theorem 5]) Let S be the center of the circle determined by the vertices B, C of .6.ABC and the excenter Ia in the interior of LA.

If

the circle S passes through the circumcenter 0, then we have the following

( 1) The circle passes through the orthocenter H, (2) HI = IO,

(3) Hla

=

1a O (Fig. 5).

Theorem 6. ( [5; Theorem 6]) Let S be the center of the circle determined by the vertices B, C and the orthocenter H of MBC.

If

the circle S passes the excenter Ia in the interior of LA, then we have the following

( 1) The circle S passes through the incenter I, (2) The circle S passes through the circumcenter 0, (3) HI = IO,

(4) Hia = Ia O (Fig. 6).

(6)

Theorem 7. ( [5; Theorem 7]) Let S be the center of the circle determined by the vertices B, C and the orthocenter H of /;:,.ABC.

If

the circle S passes through the excenter lb in the interior of LB, then we have the following :

(1) The circle S passes through the circumcenter 0,

(2) The circle S passes through the excenter le in the interior of LC, (3) The line-segment Ible is the diameter of the circle S,

(4) Hlb

=

^{lb 0,}

(5) H/c

=

^Jc^0,

(6) The lines OH and Ible are perpendicular to each other (Fig. 7).

Proof of 'rheorem 7. Note that LA

>

90 °. We have LA

=

120 °, because LBHC

=

LBibC. The proof proceeds similarly to that of Theorem 2.

i) .6.SOB is an equilateral triangle.

Hence the circumcircle O of MBC passes through the point S.

ii) The quadrangle SBOC is a rhombus.

Hence the diagonals SO and BS bisect each other and intersect at the right angles.

iii) The quadrangle HAOS is a rhombus.

iv) LlbSH

=

LlbSO.

(1) follows from i).

· (2) We have LBibC = LBicC = 60 °. Since both points lb, le are in the same side of the line BC, the sircle S passes through the point le by the converse of the Theorem of inscribed angles.

(3) follows from L lb Blc

=

90° .

--- ^---

(4) By iv) Hlb

=

hO; that is, Hlb

=

Ibo.

(5) The proof is similar to that of ( 4) . ( 6) follows from iii) .

D

Theorem 8. ( [5; Theorem 8]) Let S be the center of the circle determined by the vertices B, C and the circumcenter O of .6.ABC.

If

the circle S passes through the excenter lb in the interior of LB, then we have the following :

(1) The circle S passes through the orthocenter H,

(2) The circle S passes through the excenter le in the interior of LC (Fig. 8).

The proofs of the rest of the theorems are similar to the above, which are omitted.

(7)

References

[1] H. S. M. Coxeter : Introduction to Geometry, 2nd ed., John Wiley and Sons Inc., New York, 1980. ( ':1 7 ~ 7 -

(il# ii ·

^{w{) :}

~1PJ~J..r,

~ 2

mi,

ajf

ri,~-. ^*JX,

1982.J

[2] H. S. M. Coxeter and S. L. Greitzer : Geometry revised, New Mathematical Library, Number 6, School Mathematics Study Group, Random House, Inc., New York, 1967.

(':1-7 A 7'-,

~-v1

^o/^{7 -}

c~~~jJ: •

^w{):

^~M~WJ..rt sMsGm.•. rnJ

lli!f mJJr1±, JllJX,

1910.J

[3] K. Fujita, A. Matsushima and H. Fukaishi : A Locus of the Orthocenter of a Triangle - Instruction in Geometry by a Moving Locus on a Computer, Mem. Fae. Educ., Kagawa University II, 55 (2005), 1-13.

[ 4] K. Fujita, A. Matsushima and H. Fukaishi : A Triangle with Three Distinguished Collinear Points - Instruction of Geometry by Use of a Drawing Game on a Display, Mem. Fae. Educ., Kagawa University II, 55 (2005), 25-41.

[5] K. Fujita, A. Matsushima and H. Fukaishi : A Triangle with Distinguished Concyclic Points - Instruction of Geometry by Use of a Drawing Game on a Display, Mem.

Fae. Educ., Kagawa University II, 56(2006), 1-25. ·

[6] E. E. Moise : Elementary Geometry, From an Advanced Standpoint, Addison-Wesley Publ. Co., Reading Massachusetts, 1963.

[7] D. Pedoe : Geometry, Dover Puhl. Inc., New York, 1970.

Hiroo FUKAISHI

Department of Mathematics, Faculty of Education, Kagawa University 1-1 Saiwai-cho, Takamatsu-shi, Kagawa, 760-8522, JAPAN

E-mail address : fukaishi@ed.kagawa-u.ac.jp

Kazunori FUJITA

Department of Mathematics, Faculty of Education, Kagawa University 1-1 Saiwai-cho, Takamatsu-shi, Kagawa, 760-8522, JAPAN

E-mail address : fujita@ed.kagawa-u.ac.jp

Akiko MATSUSHIMA

Student of Master Course, Graduate School of Education, Kagawa University 1-1 Saiwai-cho, Takamatsu-shi, Kagawa, 760-8522, JAPAN

(8)

Fig.1

0

Fig. 3

.. .··

s

Fig. 2

s

Fig. 4

(9)

/A .... •···

···

Fig. 5 Fig. 6

.. •·

Fig. 7 Fig. 8

Alternative Geometric Proofs of Theorems for Con cyclic Points for a Triangle