Geometry with Trigonometry

Sec. 11.6 Mobile coordinates 215

11.6.6 Euler line of a triangle .......................

With the notationZ 7 ,Z 16 ,Z 11 for the centroid, circumcentre and orthocentre, respec-
tively, of a triangle[Z 1 ,Z 2 ,Z 3 ]we have the formulae

z 7 −z 2 =^13 (p 1 + 1 +q 1 ı)(z 3 −z 2 ),

z 16 −z 2 =

1

2

(

1 +

p^21 −p 1 +q^21

q 1

ı

)

(z 3 −z 2 ),

z 11 −z 2 =p 1

(

1 −

p 1 − 1

q 1

ı

)

(z 3 −z 2 ).

It is straightforward to check that

2

3 z^16 +

1

3 z^11 =z^2 +

2

3 (z^16 −z^2 )+

1

3 (z^11 −z^2 )=z^7 ,

and soZ 7 ∈Z 16 Z 11.
Thus we have shown thatthe centroid, circumcentre and orthocentre of any
triangle are collinear. This is a result due to Euler, after whom this line of collinear-
ity is named theEuler lineof the triangle.

11.6.7 Similar triangles ............................

For any two triangles[Z 1 ,Z 2 ,Z 3 ]and[Z′ 1 ,Z′ 2 ,Z′ 3 ]suppose that we have z 1 =z 2 +(p 1 +
q 1 ı)(z 3 −z 2 ),z′ 1 =z′ 2 +(p′ 1 +q′ 1 ı)(z′ 3 −z′ 2 ),where(p 1 ,q 1 ),(p′ 1 ,q′ 1 )are couples of real
numbers, neither being( 0 , 0 ).Then these triangles are similar in the correspondence
(Z 1 ,Z 2 ,Z 3 )→(Z′ 1 ,Z′ 2 ,Z 3 ′)if and only if p′ 1 =p 1 ,q′ 1 =±q 1.
Proof. First suppose thatp′ 1 =p 1 ,q′ 1 =q 1 so thatz 1 =z 2 +(p 1 +q 1 ı)(z 3 −
z 2 ),z′ 1 =z′ 2 +(p′ 1 +q′ 1 ı)(z′ 4 −z′ 2 ). Then we have that

|FZ 3 Z 2 Z 1 |◦=|FZ 3 ′Z 2 ′Z 1 ′|◦,|FZ 1 Z 3 Z 2 |◦=|FZ 1 ′Z 3 ′Z 2 ′|◦.

It follows by 5.3.2 that the measures of the corresponding angles of these triangles
are equal, and so the triangles are similar, with the lengths of corresponding sides
proportional. Moreover, the triples(Z 1 ,Z 2 ,Z 3 )and(Z′ 1 ,Z′ 2 ,Z 3 ′)are similarly oriented.

Z (^1)

Z 2

Z 3

W 1

Figure 11.14. Similar triangles.

Z 1 ′

Z 2

Z′ 3

W 1 ′