Geometry with Trigonometry

Sec. 11.7 Some well-known theorems 229

For the nine-point circle we take the pointZ 26 such that

z 26 =z′ 16 +

r 1

|z 15 −z′ 16 |

(z′ 16 −z 15 ),

ThenZ 26 is a point of the lineZ 16 ′Z 15 which lies on the nine-points circle and the

corresponding point

z 27 =z′ 16 −

r 1

|z 15 −z′ 16 |

(z′ 16 −z 15 ),

also lies on the nine-point circle for which[z 26 ,z 27 ]is a diameter. But in fact the

pointsZ 24 andZ 26 coincide and we then have the situation thatZ 27 ∈[Z 24 ,Z 25 ].

Thus we have that the incircle has diameter[Z 24 ,Z 25 ]and the nine-point circle has

diameter[z 24 ,z 27 ]and[z 24 ,z 27 ]⊂[z 24 ,z 25 ]. It follows that the nine-point circle lies

inside or on the the incircle. But the incircle lies inside or on the triangle[Z 1 Z 2 Z 3 ]

and it follows that the nine-points circle lies inside or on the triangle. But this yields

a contradiction as the nine-points circle passes through the mid-pointsZ 4 ,Z 5 ,Z 6 of

the sides of this triangle.

There remains the case whenr 2 =r 1 and now we must haveZ 16 ′ =Z 15 and the

nine-points circle must coincide with the incircle. In this case the two circles do not

meet in a unique point.

This result is known as the basic form ofFeuerbach’s theorem.

11.7.2 The Wallace-Simson line, 1797 ....................

We take a triangle[Z 1 ,Z 2 ,Z 3 ]and for a pointZletW 1 ,W 2 ,W 3 be the feet of the per-

pendiculars fromZto the side-linesZ 2 Z 3 ,Z 3 Z 1 ,Z 1 Z 2 , respectively.

Z 1

Z 2

Z 3

Z

W 1

W 2

W 3

Figure 11.19(a). A right sensed duo-angle.

Z 1

Z 2

Z 3

Z

W (^2) W 1

W 3

(b). A Simson-Wallace line.

Using notation like that in 11.6.3?, we suppose thatz 1 −z 2 =(p 1 +q 1 ı)(z 3 −z 2 ),z−

z 2 =(p+qı)(z 3 −z 2 ).Thenz=z 2 +p(z 3 −z 2 )+qı(z 3 −z 2 ),andsow 1 =z 2 +p(z 3 −

z 2 ).Nextw 2 =z 3 +s(z 1 −z 3 ),forsomes∈R. Hencez−w 2 =(p− 1 +qı)(z 3 −z 2 )−