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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 984.
Let CC1 cut AA1 and circle O at H and ELet C1A1 cut A2C2 at M1. Note that C1E=C1HSince ∠(A1HC)= ∠(CDC2)= ∠ (B) => quadrilateral HDC2A2 is cyclicSo ∠(HA2C)= ∠(CC2D)Triangle HCA2 similar to triangle EAC2…. (case AA)With corresponding altitudes CA1 and AC12. So A1H/A1A2=C1E/C1C2Replace C1E=C1H in above => C1H/CC2=A1H/A1A2 or C1H/CC2 x A1A2/A1H=1 …(1)3. Apply Menelaus’s theorem in triangle HA2C2 with secant C1A1MC1H/C1C2 x MC2/MA2 x A1A2/A1H =1Replace value of ( 1) to above expression we have MC2/MA2=1So M is the midpoint of A2C2
See below for the sketch of problem 984http://s25.postimg.org/b91r2quv3/pro_984_1.pngPeter Tran
Let A1C1 meet A2C2 at M'. Using Law of sine, M'C2=sin(<CC1A1)*C1C2/sin(<C1M'C2)and M'A2=sin(<A2A1M')*A1A2/sin(<A1M'A2)Evaluate M'C2/M'A2=(sin(<CC1A1)*C1C2)/(sin(<A2A1M')*A1A2),but sin(<CC1A1)/sin(<A2A1M')=A1C/AC1, which simplifies M'C2/M'A2 to (A1C*C1C2)/(AC1*A1A2)=A1C/A1A2*C1C2/AC1=1 because triangles AC1C2 and CA1A2 are similar.This proves M' to be midpoint of A2C2