Online Geometry theorems, problems, solutions, and related topics.
Proposed ProblemClick the figure below to see the complete problem 318.See more:Complete Problem 318Level: High School, SAT Prep, College geometry
Consider triangle ABC with B1H(A)H(C) as the transversal. We can say, by Menelaus's Theorem, that: [CB1/B1A)(AH(C)/H(C)B)(BH(A)/H(A)C]= -1 or AB1/B1C = - [H(C)B*H(A)C]/[AH(C)*BH(A)] ---(1)Likewise, AC1/C1B=-[H(A)C*H(B)A]/[BH(A)*CH(B)] ---(2) and BA1/A1C=-[H(B)A*H(C)B]/[CH(B)*AH(C)] ---(3). Multiplying the LHS and RHS of the three equations together, we obtain: (CB1/B1A)(AC1/C1B)(BA1/A1C) = -[(AH(B)/H(B)C)*(CH(A)/H(A)B)*(BH(C)H(C)A)]^2. However, [(AH(B)/H(B)C)*(CH(A)/H(A)B)*(BH(C)H(C)A)] = 1 by Ceva's Theorem. Therefore,(CB1/B1A)*(AC1/C1B)*(BA1/A1C) = - 1 and hence by the converse of Menelaus's Theorem A1, C1 & B1 are collinear. QED. Vihaan: email@example.com
Apply Desargues theorem on ABC and its orthic triangle to get the result.