Geometry Problem

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

Click the figure below to see the complete problem 929.

## Thursday, September 26, 2013

### Geometry Problem 929: Triangle, Orthocenter, Circumcircle, Reflection, Concurrency

Labels:
circumcircle,
concurrent,
orthocenter,
reflection,
triangle

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http://img43.imageshack.us/img43/9618/h2bb.png

ReplyDeleteLet BH,AH and CH cut the circle at B’, A’ and C’

Note that A’, B’ and C’ is the reflection of H over BC, AC and AB ..( property of orthocenter)

So F, B’, B1 are collinear , similarly for E, A’, A1 and D, C’, C1

This problem become problem 733 . see below for solution

http://gogeometry.blogspot.com/2012/02/problem-733-triangle-orthocenter.html

Let A', B', and C' be reflections of H over BC, AC, and AB respectively. By property of orthocenter, these points are on circumcircle.

ReplyDeleteLet FB' and DC' meet at P.

<FPD=180-2(<DFA+<FDA)=180-2<BAC=arcB'C'/2, making P on circumcircle.

Let DC' and EA' meet at P'.

<DP'E=180-(180-2<BDE)-(180-2<BED)=2(<BDE+<BED)-180=2(180-<ABC)-180=180-2<ABC

=arcA'C'/2, making P' on circumcircle.

P and P' are both on DC' and circumcircle, so they must coincide.