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

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

Click the figure below to see the complete problem 875.

## Tuesday, May 14, 2013

### Problem 875: Congruent Circles, Intersecting Circles, Secant, Center, Angle Trisection, Triple Angle, Triangle

Labels:
angle,
center,
circle,
congruence,
intersecting circles,
secant,
triangle,
triple,
trisection

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build BC

ReplyDeleteBC=AC

CBA=A=a

BCE=2a

BC=BE

E=BCE=2a

EBF=E+A=3a

q.e.d

Join BC. Then AC = BC = BE.

ReplyDeleteβ = α + ∠BEA

= α + ∠BCE

= α + α + ∠CBA

= α + α + α

= 3α

Draw a line from point B to C.

ReplyDeleteTriangle ABC is an isosceles triangle, AC = BC, then ∠CAB = ∠CBA.

∠BCE = 2α.

in triangle ABE if ∠AEB is x then ∠EBF is exterior angle, then β = α + x

∠BCE = ∠AEB = 2α then β = α + 2α = 3α (q.e.d)