OK Geometry Report : Task

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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