An Application of Sondat’s Theorem Regarding the Orthohomological Triangles
Ion Pătraşcu, “Fraţii Buzeşti” National College, Craiova, Romania; Florentin Smarandache, ENSIETA, Brest, France. In this article we prove the Sodat’s theorem regarding the orthohomological triangle and then we use this theorem and Smarandache-Patrascu’s theorem in order to obtain another theorem regarding the orthohomological triangles. Theorem (P. Sondat) Consider the orthohomological triangles ABC , A1 B1C1 . We note Q , Q1 their orthological centers, P the homology center and d their homological axes. The points P , Q , Q1 belong to a line that is perpendicular on d P
Q A1
B1
C1 C’
B’ C B A
A’
d
Proof. Let Q the orthologic center of the ABC the A1 B1C1 (the intersection of the perpendiculars constructed from A1 , B1 , C1 respectively on BC , CA, AB ), and Q1 the other orthologic center of the given triangle. We note { B '} = CA ∩ C1 A1 , { A '} = BC ∩ B1C1 , {C '} = AB ∩ A1 B1 . We will prove that PQ ⊥ d which is equivalent to B ' P2 − B ' Q2 = C ' P2 − C ' Q2 (1) We have that PA1 = α A1 A, PB1 = β B1 B, PC1 = γ C1C From Menelaus’ theorem applied in the triangle PAC relative to the transversals B ', C1 , A1 we obtain that B 'C α (2) = B'A γ The Stewart’s theorem applied in the triangle PAB ' implies that (3) PA2 ⋅ CB '+ PB '2 ⋅ AC − PC 2 ⋅ AB ' = AC ⋅ CB '⋅ AB ' Taking into account (2), we obtain: γ PC 2 − α PA2 = ( γ − α ) PB '2 − α B ' A2 + γ B ' C 2 (4) Similarly, we obtain:
γ QC 2 − α QA2 = ( γ − α ) QB '2 + γ B ' C 2 − α B ' A2
(5)
Subtracting the relations (4) and (5) and using the notations: PA2 − QA2 = u, PB 2 − QB 2 = v, PC 2 − QC 2 = t we obtain: γ t −αu (6) PB '2 − QB '2 = γ −α The Menelaus’ theorem applied in the triangle PAB for the transversal C ', B, A1 gives C 'B α (7) = C'A β From the Stewart’s theorem applied in the triangle PC ' A and the relation (7) we obtain: α PA2 − β PB 2 = (α − β ) C ' P 2 + α C ' A2 − β C ' B 2 (8) Similarly, we obtain:
α QA2 − β QB 2 = (α − β ) C ' Q 2 + α C ' A2 − β C ' B 2
(9)
From (8) and (9) it results
αu − β v α −β The relation (1) is equivalent to: C ' P2 − C ' Q2 =
(10)
αβ ( u − v ) + βγ ( v − t ) + γα ( t − u ) = 0
(11)
To prove relation (11) we will apply first the Stewart theorem in the triangle CAP , and we obtain: (12) CA2 ⋅ PA1 + PC 2 ⋅ A1 A − CA12 ⋅ PA = PA1 ⋅ A1 A ⋅ PA Taking into account the previous notations, we obtain: α CA2 + PC 2 − CA12 (1 + α ) = PA12 + α A1 A2 (13) Similarly, we find:
α BA2 + PB 2 − BA12 (1 + α ) = PA12 + α A1 A2
From the relations (13) and (14) we obtain: α BA2 − α CA2 + PB 2 − PC 2 − (1 + α ) ( BA12 − CA12 ) = 0
(14) (15)
Because A1Q ⊥ BC , we have that BA12 − CA12 = QB 2 − QC 2 , which substituted in relation (15) gives: t −v BA2 − CA2 + QC 2 − QB 2 = (16)
α
Similarly, we obtain the relations: CB 2 − AB 2 + QA2 − QC 2 =
u −t
AC 2 − BC 2 + QB 2 − QA2 =
β v−u
γ By adding the relations (16), (17) and (18) side by side, we obtain t −v
+
u −t
+
(17) (18)
v −u
(19) =0 α β γ The relations (19) and (11) are equivalent, and therefore, PQ ⊥ d , which proves the Sondat’s theorem. In [2] it was proved the Smarandache-Patrascu Theorem: If the triangles ABC and A1B1C1 inscribed into the triangle ABC are orthohomological, where Q is the center of orthology (i.e. the point of intersection of the perpendiculars in A1 on BC, in B1 on AC, and in C1 on AB), and Q1 is the second center of orthology of the triangles ABC and A1B1C1, and A2B2C2 is the pedal triangle of Q1, then the triangles ABC and A2B2C2 are orthohomological. Now we prove another theorem: Theorem (Pătraşcu-Smarandache) Consider the triangle ABC and the inscribed orthohomological triangle A1 B1C1 , with Q , Q1 their centers of orthology, P the homology center and d their homology axes. If A2 B2C2 is the pedal triangle of Q1 , P1 is the homology center of triangles ABC and A2 B2C2 , and d1 their homology axes, then the points P, Q, Q1 , P1 are collinear and the lines d and d1 are parallel. Proof.
Applying the Sondat窶冱 theorem to the ortho-homological triangle ABC and A1 B1C1 , it results that the points P, Q, Q1 are collinear and their line is perpendicular on d . The same theorem applied to triangles ABC and A2 B2C2 shows the collinearity of the points P1 , Q, Q1 , and the conclusion that their line is perpendicular on d1 . From these conclusions we obtain that the points P, Q, Q1 , P1 are collinear and the parallelism of the lines d and d1 . References [1] [2]
Cﾄフﾄネin Barbu, Teoreme Fundamentale din Geometria Triunghiului, Editura Unique, Bacﾄブ, Romania, 2008. Florentin Smarandache, Multispace & Multistructure Neutro-sophic Transdisciplinarity (100 Collected Papers of Sciences), Vol. IV. North-European Scientific Publishers, Hanko, Finland, 2010.