A SYNTHETIC PROOF OF DAO S GENERALIZATION OF GOORMAGHTIGH S THEOREM

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Global Journal of Advanced Research on Classical and Modern Geometries ISSN: 84-5569, Vol3, (014), Issue, pp15-19 A SYNTHETIC PROOF OF DAO S GENERALIZATION OF GOORMAGHTIGH S THEOREM TRAN HOANG SON

1930, René Goormaghtigh, French engineer and geometrician, expanded Droz-Farny theorem [1,, 3] with a nice theorem as follow Theorem 11 (Goormaghtigh [4]) Given triangle ABC and point P distinct from

1 Global Journal of Advanced Research on Classical and Modern Geometries ISSN: , Vol3, (014), Issue, pp15-19 A SYNTHETIC PROOF OF DAO S GENERALIZATION OF GOORMAGHTIGH S THEOREM TRAN HOANG SON ABSTRACT Using the concept of cross ratio, we give a synthetic proof of Dao s generalization of Goormaghtigh s theorem MSC 010: 51M04, 51M5 Keywords: cross ratio, signed distances 1 INTRODUCTION In 1930, René Goormaghtigh, French engineer and geometrician, expanded Droz-Farny theorem [1,, 3] with a nice theorem as follow Theorem 11 (Goormaghtigh [4]) Given triangle ABC and point P distinct from A, B, C A line passes through P A 1, B 1, C 1 belong to BC, CA, AB respectively such that PA 1, PB 1, PC 1 are the images of PA, PB, PC respetively by reflection R Then, A 1, B 1, C 1 are collinear Notation R refers to reflection against Theorem 11 is called Goormaghtigh s theorem [4] When P is the orthocenter of triangle ABC, theorem 11 actually becomes Droz-Farny theorem Proof of theorem 11 can be found in [5, 6] In 014, OTDao expanded theorem 11 with two theorems [7] In this article, we are first going to expand OTDao s second theorem with theorem 1 and more beautifully restate OTDao s first theorem with theorem 13 Then, we are going to prove theorems 1 and 13 Please note that, in terms of ideas, the way we prove theorems 1 and 13 is completely different from the way that theorem 11 is proved in [1] and [] Theorem 1 Given triangle ABC and point P distinct from A, B, C A line passes through P α is any real number Let A 1, B 1, C 1 belong to BC, CA, AB respectively such that PA 1, PB 1, PC 1 are the images of PA, PB, PC respectively by transformation R α P R Then, A 1, B 1, C 1 are collinear Notation R α P refers rotation around P with angle of rotation α When α = 0, theorem 1 becomes theorem 11 When α = π, theorem 1 becomes OTDao s second theorem Theorem 13 (Dao [7]) Given triangle ABC and point P distinct from A, B, C Lines and cut at P Points A 1, B 1, C 1 belong to BC, CA, AB respectively such that (PA, PA 1,, ) = (PB, PB 1,, ) = (PC, PC 1,, ) = 1 Then, A 1, B 1, C 1 are collinear 15

2 Tran Hoang Son When, theorem 13 becomes theorem 11 Theorem 13 is a different, more interesting reiteration of OTDao s first theorem Before we prove theorems 1 and 13, note that notation AB refers to the signed length from point A to point B PROOF OF THEOREM 1 There are two cases to consider Case 1 α 0 (modπ) Then, R α P R = R Ignore platitudinous situations: passes through a vertex of triangle ABC; passes through two vertices of triangle of ABC Let A, B be the intersections of PC 1 and BC, CA respectively (see f1) Since reflection preserves cross ratio, Figure 1 : A B = (BCA A 1 C A C 1A ) = P(BCA 1 A ) = P(BCA 1 C 1 ) = P(B 1 C 1 AC) = P(B 1 B AC) = P(ACB 1 B ) = (ACB 1 B ) = B 1A : B A B 1 C B C From this, noting that A, B, C 1 are collinear, by Menelaus theorem, we have A 1 C B 1C B 1 A C 1A C 1 C = A B A C B C B A C 1A C 1 C = 1 Hence, by Menelaus theorem, A 1, B 1, C 1 are collinear Case α 0 (modπ) Let line pass through P such that (, ) α (modπ) Apparently, R α P R = (R R ) R = R (R R ) = R id = R From this, noting that P belongs to, according to case 1, we can deduce that A 1, B 1, C 1 are collinear We need two lemmas 3 PROOF OF THEOREM 13 Lemma 31 If BC, CA, AB are parallel to B 1 C 1, C 1 A 1, 1 respectively, then two triangles ABC and 1 C 1 are similar in the same direction 16

3 A synthetic proof of Dao s generalization of Goormaghtigh s theorem Proof We have BC // B 1 C 1 ; CA // C 1 A 1 ; AB // 1 Therefore, (BA, BC) (B 1 A 1, B 1 C 1 ) (modπ) and (CA, CB) (C 1 A 1, C 1 B 1 ) (modπ) Hence, triangles ABC and A B C are similar in the same direction Lemma 3 Given two triangles ABC and 1 C 1 which are similar in the same direction A, B, C are the midpoints of AA 1, BB 1, CC 1 respectively Then, triangle A B C are similar to triangles ABC and 1 C 1 in the same direction Proof Let M, N be the midpoints of AB 1, AC 1 respectively (see f) Because M, N, A are the midpoints of AB 1, AC 1, AA 1 respectively, MN, NA, A M are parallel to B 1 C 1, C 1 A 1, 1 respectively Therefore, by lemma 31, triangles A MN and 1 C 1 are similar in the same direction (1) As A, B, C, M, N are the midpoints of AA 1, BB 1, CC 1, AB 1, AC 1 respectively, MA = 1 B 1 A 1 ; MB = 1 AB; NA = 1 C 1 A 1 ; NC = AC Figure From this, noting that triangles ABC and 1 C 1 are similar in the same direction, ( MA, MB ) ( B 1 A 1, AB) π + ( B 1 A 1, BA) (modπ) π + ( C 1 A 1, CA) ( C 1 A 1, AC) ( NA, NC )(modπ) MA = B 1A 1 MB AB = B 1A 1 BA = C 1A 1 CA = C 1A 1 AC = NA NC Thus, triangles A MB and A NC are similar in the same direction Therefore, triangles A MN and A B C are similar in the same direction () From (1) and (), deduce that 1 C 1 and A B C are similar in the same direction In other words, triangle A B C are similar to triangles ABC and 1 C 1 in the same direction Return to the proof of theorem 13 Let A, B be the intersections of PC 1 and BC, CA respectively Let A 3, B 3, C 3 be the intersections of PA 1, PB 1, PC 1 and the lines parallel to, passing through A, B, C respectively Let A 0, B 0, C 0 be the intersections of and AA 3, BB 3, CC 3 respectively (see f3) 17

4 Tran Hoang Son Figure 3 Because (PA, PA 1,, ) = (PB, PB 1,, ) = (PC, PC 1,, ) = 1, (PA, PA 3, PA 0, ) = (PB, PB 3, PB 0, ) = (PC, PC 3, PC 0, ) = 1 Then, combined with the fact that AA 3, BB 3, CC 3 are all parallel to, we can deduce that A 0, B 0, C 0 are the midpoints of AA 3, BB 3, CC 3 respectively If BC, CA, AB are parallel to B 3 C 3, C 3 A 3, A 3 B 3 respectively, then by lemma 31, triangles ABC and A 3 B 3 C 3 are similar in the same direction From this, noting that A 0, B 0, C 0 are the midpoints of AA 3, BB 3, CC 3 respectively, by lemma 3, we can deduce that A 0, B 0, C 0 are not collinear, contradiction Thus, BC, CA, AB are not respectively parallel to B 3 C 3, C 3 A 3, A 3 B 3 Without the loss of generality, assume that BC and B 3 C 3 are not parallel Let S be the intersection of BC and B 3 C 3 Let A, A 4 be the intersections of BC and C 1 P, AA 3 respectively Let A 5, A 6, A 7 be the intersections of B 3 C 3 and AP, CP, AA 3 respectively Apparently, S belongs to Applying Cevas theorem to triangle SC 3 C, noting that SC 0, C 3 A, CA 6 are concurrent (at P), we have C 0 C 3 C 0 C A C A S A 6S A 6 C 3 = 1 Combined with the fact that C 0 is the midpoint of C 3 C, we have A S A C = A 6S A 6 C 3 Therefore, by Thales theorem, A A 6 // CC 3 (3) 18

5 A synthetic proof of Dao s generalization of Goormaghtigh s theorem Applying Menelaus theorem to triangles A 0 SA 4 and A 0 SA 7, noting that A 1, A 3, P are collinear and A 5, A, P are collinear, we have A 1 S A 3A 4 PA 0 A 1 A 4 A 3 A 0 PS = 1 = A 5S AA 7 PA 0 A 5 A 7 AA 0 PS From this, noting that A 0 is the midpoint of both AA 3 and A 4 A 7, deduce that A 1 S = A 5S A 1 A 4 A 5 A 7 Therefore, by Thales theorem, A 1 A 5 // A 4 A 7 (4) From (3) and (4), deduce that BB 3 //CC 3 //A 1 A 5 //A A 6 Hence, A 1 C : A B A C = (BCA 1A ) = (B 3 C 3 A 5 A 6 ) = P(B 3 C 3 A 5 A 6 ) = P(B 1 B AC) = P(ACB 1 B ) = (ACB 1 B ) = B 1A B 1 C : B A B C From this, noting that A, B, C 1 are collinear, by Menelaus theorem, deduce that A 1 C B 1C B 1 A C 1A C 1 B = A B A C B C B A C 1A C 1 B = 1 Thus, by Menelaus theorem, A 1, B 1, C 1 are collinear 4 ACKNOWLEDGMENT The author thanks Professor Nguyen Minh Ha for his invaluable ideas in helping me to complete this article REFERENCES [1] A Droz-Farny, Question 1411, Ed Times 71 (1899) pp [] J L Ayme, A synthetic proof of the Droz-Farny line theorem, Forum Geom, Vol 4, (004) pp 19-4 [3] N M Ha and L T Vinh, Purely synthetic proof of the Generalized Droz-Farny theorem, Global Journal of Advanced Research on Classical and Modern Geometries, Vol 1, Issue, (01), pp [4] R Goormaghtigh, Sur une généralisation du théoreme de Noyer, Droz-Farny et Neuberg, Mathesis 44 (1930) 5 [5] Darij Grinberg, Message Hyacinthos 8664, November/9th/003 [6] T Andreescu and CPohoata, Black Euclidean geometry: Droz-Farny demystified, Mathematical Reflections 3 (01) pp -3 [7] OTDao, Message Advanced Plane Geometry 171, April/6th/014 UNIVERSITY OF EDUCATION, HANOI, VIETNAM address: transonsp97@gmailcom 19

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