See complete Problem 138

Nagel's Theorem, Orthic Triangle, Altitudes, Circumradius, Perpendicular. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Tuesday, July 22, 2008

### Elearn Geometry Problem 138 Nagel's Theorem

Labels:
altitude,
angle,
circumcircle,
circumradius,
Nagel theorem,
orthic triangle,
perpendicular,
triangle

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as O is the circumcenter,

ReplyDelete∠OBC=90°-∠CAB=∠ABE.

as FACD concyclic,

∠BDF=∠CAB.

thus BO⊥DF is obvious, so are the others.

angleBAO=angleABO=90-C and angleAHE=angleAFE(H is ortho centre and AFHE is cyclic) but angle AHE=90-(90-C)=C so angleAFE=C hence AO is perpendicular to EF simillarly remaining we can prove

ReplyDeleteThe acute angle between the tangent at B to the circumcircle and the chord BA equals the angle in the alternate segment angle BCA which in turn equals angle DFB (ACDF being a cyclic quadrilateral).

ReplyDeleteTherefore FD is parallel to the tangent at B and hence perpendicular to the radius BO.

About problem 138.

ReplyDeleteIn the solution by Vijay (March 20, 2010), I didn't understand how the fact that angleAFE=C leads to the conclusion that AO is perpendicular do EF.

Can someone help me?

Form 1:

ReplyDeleteO and H are isogonal conjugate, so the sides of pedal triangle of H are orthogonal to the respective line AO, BO or CO and reciprocally.

Form 2:

We know that A, B and C are the excenters of tr DEF, so by the well known existence of Bevan's point, we get that perpendiculars through A, B and C to the respective side of tr DEF are concurrent.

Form 3:

tr DEF is orthologic to tr ABC by definition, so by the orthology theorem, tr ABC must be orthologic to tr DEF.