Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 968.
Let z(P) be the complex number representing P. Let z(A)=0, z(F)=a, z(C)=a-biThenz(B)=(1/2 a + √3/2 b)+(√3/2 a - 1/2 b)iz(E)=a+(√3/2 a - 1/2 b)iz(M)=(1/4 a + √3/4 b)+(√3/4 a - 1/4 b)iEasy to check that z(F) + (-1/2 + √3/2 i) z(E) + (-1/2 - √3/2 i) z(M) = 0Hence, EMF is equilateral triangle.
Connect CM Note that CM ⊥BM => MBEC is cocyclicSince M is midpoint of AB => triangle EMF is isoscelesIn circle MBEC , ∠MEC=∠MBC=60 => EMF is equilateral
CM is perpendicular to AB hence MBEC is con cyclic, so < MEC = < MBC = 60Also ACFM is con cyclic so < EFM = < MAC = 60So 2 angles of Tr. FEM have been proved to be 60 hence this triangle must be equilateral Sumith PeirisMoratuwaSri Lanka