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

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

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## Saturday, September 30, 2017

### Geometry Problem 1348: Three Concentric Circles, Equilateral Triangle, Perpendicular Bisector, Congruence

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Drop perpendiculars from O to EA & B to DA

ReplyDeleteSince BA=OA => m(BAD) = m(OAE)

Hence triangles BAD & OAE are congruent

=> DA = EA

m(DAE)=m(DAO)+m(OAE)

=>m(DAE)=60-m(BAD)+m(OAE)

=>m(DAE)=60

Hence DAE is equilateral

To Anonymous: Please explain your second line. Thanks

DeleteIt is just construction of an equilateral triangle with vertices onto 3 concentric circles

Delete

ReplyDeleteLet a circle (A,AD) that intersects the circle C_3 at the point Z.Then triangle DBA=triangle ZOA (AD=AZ,AB=AO,DB=ZO).So triangle DAZ is

equilateral.If ZN perpendicular of AD,then AN=ND or N,M=coincide.Therefore and

the points Z,E= coincide.So triangle DAE is equilateral .

APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE