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

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

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## Friday, July 8, 2016

### Geometry Problem 1233: Triangle, Euler Line, Orthic Axis, Perpendicular, 90 Degrees, Orthic Triangle

Labels:
Euler line,
orthic axis,
orthic triangle,
perpendicular

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https://goo.gl/photos/hm7ruYYeKsGei8867

ReplyDeleteDraw 9 points circle of triangle ABC. The center N of this circle will be on Euler line NGH.

This circle will pass through A1 , B1, C1, and midpoints of sides of triangle ABC.

Draw new circle diameter ẠC

From B2 draw a tangent line to circle O1 at D.

Since( A,C, B1, B2)= -1 => DB1 ⊥ AC

Power of B2 to circle O= B2C . B2A=BD^2

But BD^2= B2B1. B2O1 (relationship in right triangle)= power of B2 to circle N

So B2 is on the radial axis of circles N and Ọ

Similarly C2 and A2 are on radial axis of circles N and O

Radial axis A2C2B2 will perpendicular to line connected centers of circles O and N

https://goo.gl/photos/AJ3pAb2CxNFo9j26A

ReplyDeleteNew improved solution to problem 1233

Draw nine- points circle of triangle ABC. The center N of this circle will be on Euler line NGH.

This nine- points circle will pass through A1 , B1, C1, and midpoints of sides of triangle ABC.

Quadrilateral ACA1C1 is cyclic => B2C. B2A=B2A1. B2C1 (power of B2 to circle ACA1C1= power of B2 to circle O= power of B2 to circle N)

So B2 have the same power to both circles O or N => B2 will be on the radical axis of circles O and N.

Similarly ABA1B1 and BCB1C1 are cyclic and C2B.C2A=C2A1. C2B1 ,A2B.A2C= A2C1. A2B1 => C2 and A2 are on radical axis of circles N and O

Radical axis A2C2B2 will perpendicular to line connected centers of circles O and N