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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 810.
Let CD cut AB at INote that ABD, ABC and CBF are equilateral trianglesAnd G is the midpoint of BC and AFIn triangle DBC, DG and BI are medians => N is the centroid of DBCSo BN=2.IN Due to symmetric we have AM=2.IN=MN=NB
Since DE is diameter with center A, we have EA = AD. Since CADB is a rhombus, AC//DB, hence, EH = HB. Now in ΔEDB, A and H are mid-points, which means BA and DH are medians. Therefore, M is the centroid of ΔEAB. Thus, AM:MB = 1:2. Similarly, N is the centroid of ΔFAD. We also have AN:NB = 2:1. As a result, AM = MN = NB.
Typo: Therefore, M is the centroid of ΔEDB.
Observe that triangle CAG, triangle AGB, triangle FGB are congruent and they are 30-60-90 triangles. (proof is skipped)Hence AF is the perpendicular bisector of BC.Now insert coordinate system with the midpoint of AB be O(0,0), and auxillarily let B(1,0).Then D = (0,-sqrt3) ; C = (0,sqrt3) ; G is the midpoint hence = (1/2, (sqrt3)/2)Line DG : y + sqrt3 = 3*sqrt3*xWhen DG cuts AB, y = 0, x = 1/3. => N(1/3,0)Hence ON is one-thrid of OB.Symmetrically on the other side,q.e.d.
This is a proof using the result in problem 809.*(using the points in problem 809)Let the point where BC cuts DE named P.Then since BC = CE, and CE//BD,trianglePCE is congruent to trianglePBD.Hence PB=PC.So point P is acutally the point G in this problem.
First ABC is an equilateral triangle and G is the midpoint of CB, H is the midpoint of CA.Let's call P the point where HG(extended) cuts BF.. HG=GP (because CBF is an equilateral triangle) . therefore using TALETE (HG || AB) follows that MN=NB and so MN=NB=AM (by simmetry)
AB = 2AH = BD and since Tr.s AHM and BDM are similar BM = 2AM Similarly AN = 2BN and the result follows Sumith PeirisMoratuwaSri Lanka