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

Labels:
60,
circular sector,
congruence,
midpoint,
perpendicular

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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