Proposed Problem
Click the figure below to see the complete problem 365 about Circular Sector of 60 degrees, Midpoints, Perpendicular, Congruence.
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Complete Geometry Problem 365
Level: High School, SAT Prep, College geometry
Tuesday, October 13, 2009
Problem 365. Circular Sector of 60 degrees, Midpoints, Perpendicular, Congruence
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GF = 1/2 AB ( as middle line of OAB ) (1)
ReplyDeleteGF // AB
DE = 1/2 AB ( as middle line of ACB ) (2)
DE // AB
from (1) & (2)
GF = De, GF // DE
at the same way
GD = EF, GD // EF
it easy FGDE rombhus
FGDE is a parallelogram per C.t.e.o comment not rhombus.
ReplyDeleteTo Peter
ReplyDeleteIt is necessary to be rhombus
OAB equilateral => OC = AB => GD = DE = EF = GF
It is OK
ReplyDeleteMidpoints of a quadrilateral connected together always form a parallelogram. But the two diagonals of Quadrilateral ACBO are equal. (Connect OC, revealing OC=OA=OB. Connect AB, revealing Equilateral OAB, so OC=AB.) Quadrilateral ACBO is a rhombus.
ReplyDeleteTherefore, DF is perpendicular to EG.
Anonymous
ReplyDeleteThank you for your explanation
OA=OC=OB=AB since Tr. OAB is equilateral
ReplyDeleteThen from mid point theorem GDEF is a rhombus whose diagonals are perpendicular to each other
Sumith Peiris
Moratuwa
Sri Lanka