Smarandache’s Cevians Theorem (II) Edited by Dr. M. Khoshnevisan Neuro Intelligence Center, Australia E-mail: mkhoshnevisan@neurointelligence-center.org
Abstract. In this paper we present the Smarandache’s Cevians Theorem (II) in the geometry of the triangle. Smarandache’s Cevians Theorem (II) In a triangle Δ ABC we draw the Cevians AA1 , BB1 , CC1 that intersect in P . Then: PA PB PC AB ⋅ BC ⋅ CA ⋅ ⋅ = PA1 PB1 PC1 A1 B ⋅ B1C ⋅ C1 A Solution 6. In the triangle Δ ABC we apply the Ceva’s theorem: AC1 ⋅ BA1 ⋅ CB1 = − AB1 ⋅ CA1 ⋅ BC1 (1) In the triangle Δ AA1 B , cut by the transversal CC1 , we’ll apply the Menelaus’ theorem: AC1 ⋅ BC ⋅ A1 P = AP ⋅ A1C ⋅ BC1 (2) In the triangle Δ BB1C , cut by the transversal AA1 , we apply again the Menelaus’ theorem: A B1
C1 P
C B
A1
BA1 ⋅ CA ⋅ B1 P = BP ⋅ B1 A ⋅ CA1 (3) We apply one more time the Menelaus’ theorem in the triangle Δ CC1 A cut by the transversal BB1 : AB ⋅ C1 P ⋅ CB1 = AB1 ⋅ CP ⋅ C1 B (4) We divide each relation (2), (3), and (4) by relation (1), and we obtain: PA BC B1 A (5) = ⋅ PA1 BA1 B1C