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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 824.
Apply Menelaus’s theorem to triangle OAB and transversal DB’A’ we have(DA/DB).(B’B/B’O).(A’O/A’A)=-1Apply Menelaus’s theorem to triangle OBC and transversal FC’B’ we have(FB/FC).(C’C/C’O).(B’O/B’B)=-1Apply Menelaus’s theorem to triangle OAC and transversal EC’A’ we have(EC/EA).(A’A/A’O).(C’O/C’C)=-1Multiplying the three equalities we obtain after simplification.(DA/DB).(FB/FC).(EC/EA)=-1 So D,E,F are collinear per converse of Menelaus’s theorem
Using homogeneous coordinates in projective geometry. WLOG, we may let A=[1,0,0], B=[0,1,0], C=[0,0,1] and O=[1,1,1]. Then we haveline OA: y=zline OB: x=zline OC: x=yThus we may let A'=[a,1,1], B'=[1,b,1], C'=[1,1,c]. Now we haveline AB: z=0line A'B': (1−b)x−(a-1)y+(ab−1)z=0∴ D = AB∩A'B' = [a−1,1−b,0]line AC: y=0line A'C': (c−1)x−(ac-1)y+(a−1)z=0∴ E = AC∩A'C' = [a−1,0,1−c]line BC: x=0line B'C': (bc−1)x−(c-1)y+(1−b)z=0∴ F = BC∩B'C' = [0,b−1,1−c]Now consider the determinant|a−1 1−b 0 ||a−1 0 1−c| = −(a−1)(b−1)(1−c)−(a−1)(1−b)(1−c) = 0| 0 b−1 1−c|Hence, D,E,F are collinear.
Another proof using projective geometry but without coordinates. Since O lies on AA', we can write O=mA+nA'. Similarly, we have O=pB+qB' and O=rC+sC'. So we havemA+nA' = pB+qB' ⇒ mA−pB = −nA'+qB' = DmA+nA' = rC+sC' ⇒ mA−rC = −nA'+sC' = EpB+qB' = rC+sC' ⇒ pB−rC = −qB'+sC' = FD+E+F = (mA−pB)+(mA−rC)+(pB−rC) = 0So D,E,F are collinear.