Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

## Wednesday, August 29, 2018

### Geometry Problem 1382: Circle, Diameter, Radius, Perpendicular, Congruence

Labels:
circle,
congruence,
diameter,
geometry problem,
perpendicular,
radius

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Angle COG = Angle FPE = 180 - Angle BOC

ReplyDeleteE, O, F, P, are on the circle with center O1 and diameter OP

Connect F with O1, and this line intersects the circle with center O1 at L.

FL = OP = OC

Angle FLE = Angle FPE

triangle OCG = triangle FLE

CG = EF

Erina NJ

Good work Erina

DeleteThank You!

Deletehttps://photos.app.goo.gl/HnS668zUQ5v6RaBC9

ReplyDeleteNote that quadr. OEPF is cyclic

So ∠ (FOA)= ∠ (FPE)= u

Draw PN ⊥ EF , N is the intersection of FN to circle OEPF

Note that NE and OP are diameters of circle OEPF

And ∠ (FNE)= ∠ (FPE)= u

Triangle OCG congruent to ENF ( case HA) so CG= EF

Thank you for the picture of the geometry problem, and the comment you made to my solution.

DeleteNote that F,O,E,P concylic, so by Sine rule, EF = OPsin<P = OCsin<COG = CG. done

ReplyDeleteAlex Toronto