## Tuesday, October 13, 2009

### Problem 365. Circular Sector of 60 degrees, Midpoints, Perpendicular, Congruence

Proposed Problem
Click the figure below to see the complete problem 365 about Circular Sector of 60 degrees, Midpoints, Perpendicular, Congruence. See more:
Complete Geometry Problem 365
Level: High School, SAT Prep, College geometry

1. GF = 1/2 AB ( as middle line of OAB ) (1)
GF // 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

2. FGDE is a parallelogram per C.t.e.o comment not rhombus.

3. To Peter
It is necessary to be rhombus
OAB equilateral => OC = AB => GD = DE = EF = GF

4. 5. Midpoints 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.
Therefore, DF is perpendicular to EG.

6. Anonymous
Thank you for your explanation

7. OA=OC=OB=AB since Tr. OAB is equilateral

Then from mid point theorem GDEF is a rhombus whose diagonals are perpendicular to each other

Sumith Peiris
Moratuwa
Sri Lanka