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
Subscribe to:
Post Comments (Atom)
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