# OK Geometry Report : Task

Author:: | -
Source: | C:\Users\Zlatan\Desktop\CADGME14\NinePoint3.pro
Comment: |

## TASK

| 1 NINE POINT CIRCLE Given is a triangle △ABC, let H be its orthocentre. Let A', B', C' be the bases of the heights of ABC. Let A'',B'',C'' be the midpoints of the sides of △ABC. Finally, let A₁,B₁,C₁ be the midpoints of the segments AH, BH, CH Then the points A',B',C', A'',B'',C'', A₁, B₁,C₁ lay on a circle called nine-point circle of △ABC. Furthermore, the segments A₁A'', B₁B'', C₁C'' are diamters of the nine-point circle.

### WORKOUT

| 2 Overview of the proof | We claim in 2 that A₁B₁A''B'' is a rectangle. Thus A₁,B₁,A'',B'' lay on the same circle, and A₁A'' is its diameter. We claim in #4 that A₁C₁A''C'' is a rectangle. Thus A₁,C₁A'',C'' lay on the same circle, and A₁A'' is its diameter. The rectangles in 3 and #4 posess a common diagonal A₁A'', thus A₁,A'',B₁,B'',C₁,C'' lay on a common circle. Finally we prove in 5 that A', B', and C' lay on the above mentioned circle.
| 3.0 A''B''B₁A₁ is a rectangle. | We claim that A₁B₁ is parallel to B''A'' (#3.2). We claim that A₁B'' is parallel to B₁A'' (#3.1). We claim that A₁B₁ is perpendicuar to A₁B'' (#3.3). Thus A₁B₁A''B'' is a right angled parallelogram , i.e. a rectangle.
| 3.1 A₁B'' || B₁A''. | A₁ is the midpoint of AH. B₁ is the midpoint of BH. By Thales theorem △ABH and △A₁ B₁H are similar and AB || A₁B₁
| 3.2 A₁B₁ || A''B'' | A'' is the midpoint of BC. B'' is the midpoint of AC. By Thales theorem △ABC and △B''A''C are similar and AB || A''B''. Thus A₁B₁ || A''B''.
| 3.3 A₁B₁ ⊥ AB'' | A₁B₁ || AB by #3.2. A₁B'' || C'C by #3.1. Since CC' is the altitude to AB, CC' ⊥AB, and A₁B₁ ⊥ A₁B''.
| 4 A₁C''A''C₁ is a rectangle | This is true for the same reasons as #3.0. Note that bothe rectangles share the same diagonal A₁A''. Thus A''C₁B''A₁C''B₁ lay on the same circle.
| 5.0 B' and C' lay on the circle with diameter A₁A'' |
| 5.1 A₁A'', C₁C'', B₁B'' are diameters of a common circle |
| 5.2 B' lays on the circle A''B''C'' | ∠B''B'B₁ is a right angle. Thus B' lays on the circle that has B₁B'' as diameter, by Thales theorem.
| 5.3 A' and C' lay on circle A''B''C''. | ∠C''C'C₁ is a right angle. Thus C' lays on the circle that has C₁C'' as diameter, by Thales theorem.

## Notes