Interactive step-by-step animation using GeoGebra. Post your solution in the comment box below.

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

Details: Click on the figure below.

## Wednesday, November 13, 2019

### Dynamic Geometry Problem 1446: The Lemoine Line

Labels:
circumcircle,
collinear,
dynamic geometry,
GeoGebra,
Lemoine theorem,
tangent,
triangle

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https://photos.app.goo.gl/qcEU22yNJ3eFWNh76

ReplyDeleteLet circle center E, radius=EA cut BC at M and L ( see sketch)

Note that ∠ (ABC)= ∠ (EAC)=u and triangle EAM is isosceles

∠ (BAM)= ∠ (AMC)-u= ∠ (MAC) => AM and AL are angle bisectors of angle BAC

Calculate MC= a.k/(1+k) , MB=a/(1+k)

LC=a.k/(1-k) ; LB=a/(1-k)

And EC= ak^2/(1-k^2) ; EB= a/(1-k^2) ; EC/EB= k^2= b^2/c^2

Similarly we have DB/DA= a^2/b^2 and FA/FC= c^2/a^2

Verify that EC/EB x DB/DA x FA/FC= 1 => E,D,F are collinear per Menelaus’s theorem

Just apply Pascal on AABBCC.

ReplyDeleteSorry Peter Tran, your sketch at https://photos.app.goo.gl/qcEU22yNJ3eFWNh76 can not be open. Thanks

ReplyDelete