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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 704.
Quadrilateral BEHD is concylic . BH is the diameter of circle (EBHD).Quadrilateral AEDC is concylic . AC is the diameter of circle (AEDC). These 2 circles intersect at E and D so line connected 2 centers will perpendicular to ED.Peter Tran
AEDC is a cyclic quadrilateral, AC is the diameter and F is the circumcenter:then EF = DF.EBDH is a cyclic quadrilateral,BH is the diameter and M is the circumcenter:then EM = DM.Therefore EMDF is a Kite, then FM is perpendicular to DE.
Tr.s FEM & FDM are congruent SSS, hence FM is the perpendicular bisector of DESumith PeirisMoratuwaSri Lanka
Problem 704Is ME=MD=BH/2, FE=FD=AC/2.Then the points F, M belong to the perpendicular bisector of ED.Therefore FM is perpendicular to DE.MANOLOUDIS APOSTOLIS FROM GREECE