## Wednesday, August 29, 2018

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

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below. 1. Angle COG = Angle FPE = 180 - Angle BOC
E, 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

1. Good work Erina

2. 2. https://photos.app.goo.gl/HnS668zUQ5v6RaBC9

Note 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

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

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