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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 809.
Let CD cut AB at I and BC cut DE at JNote that ABD, ABC and CEF are equilateral trianglesWe have DB//AC//CE => DBEC is a parallelogramAnd J is the midpoint of BC In triangle DBC, DJ and BI are medians => N is the centroid of DBCSo BN=2.IN Due to symmetric we have AM=2.IN=MN=NB
Since FC//CB//AD, and FC = CB = AD, so FADC is a parallelogram, thus, FD and AC bisect each other. Join AD and DC. Consider ΔADC, DM bisects AC, and AB bisects DC. So M is the centroid of ΔADC. Let AB meets DC at H. Then AM:MH = 2:1. Similarly, HN:NB = 1:2. Since MH = HN, so AM:MN:NB = 1:1:1. As a result, AM = MN = NB.
Using vector to solve the problem.Let vector(AC) = u, vector(BC) = v.Then vector(DE) = 2u+v, vector(BA) = v-u.Now let vector(DN) = p*vector(DE) =p*(2u+v);vector(BN) = q*vector(BA) = q*(v-u).Since vector(DN) = vector(DB) + vector (BN)=> p*(2u+v) = u + q*(v-u)=> p = q = 1/3.Therefore BN = (1/3)BASymmetrical on the other side, q.e.d.
Since Tr.s AMD and BMF are similar BM = 2AM. Similarly AN = 2BN and the result follows Sumith PeirisMoratuwaSri Lanka
Problem 809Is AB=AC=AD=BD=CE=CF=BC then <ABE=90.AD intersects EB at K then <DBK=30=<DKB.So DB=DK. But ΑD=DK and EB=BK so point N is centroid of triangle AEK.Then NB=AB/3.Similar AM=AB/3.Therefore AM=MN=NB.APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE