Geometry with Trigonometry

Sec. 10.11 A case of Pascal’s theorem, 1640 177

for somet=0inR. By 9.4.1 the left-hand side is equal to

cosφ 2 −cosθ 1 +ı(sinφ 2 −sinθ 1 )

cosθ 2 −cosφ 1 +ı(sinθ 2 −sinφ 1 )

=

−2sin(^12 φ 2 +^12 θ 1 )sin(^12 φ 2 −^12 θ 1 )+ 2 ıcos(^12 φ 2 +^12 θ 1 )cos(^12 φ 2 −^12 θ 1 )

−2sin(^12 θ 2 +^12 φ 1 )sin(^12 θ 2 −^12 φ 1 )+ 2 ıcos(^12 θ 2 +^12 φ 1 )cos(^12 θ 2 −^12 φ 1 )

=

sin(^12 φ 2 −^12 θ 1 )

sin(^12 θ 2 −^12 φ 1 )

cis(^12 φ 2 +^12 θ 1 )

cis(^12 θ 2 +^12 φ 1 )

=

sin(^12 φ 2 −^12 θ 1 )

sin(^12 θ 2 −^12 φ 1 )

cis(^12 φ 2 +^12 θ 1 −^12 θ 2 −^12 φ 1 ).

Thus

cis(^12 φ 2 +^12 θ 1 −^12 θ 2 −^12 φ 1 )=

sin(^12 θ 2 −^12 φ 1 )

sin(^12 φ 2 −^12 θ 1 )

t.

But the absolute value here is 1, so the right-hand side is±1. Thus we have either

1

2 φ^2 +

1

2 θ^1 −

1

2 θ^2 −

1

2 φ^1 =^0 F,

or

1

2 φ^2 +

1

2 θ^1 −

1

2 θ^2 −

1

2 φ^1 =^180 F.

In each case, we have thatφ 2 −φ 1 =θ 2 −θ 1 and soFZ 2 OW 2 =FZ 1 OW 1.

If (Z 1 ,W 1 ), (Z 2 ,W 2 ), (Z 3 ,W 3 )
are distinct pairs of points all on
a circle and such that Z 1 W 2 ‖
W 1 Z 2 and Z 2 W 3 ‖W 2 Z 3 then
Z 1 W 3 ‖W 1 Z 3.
Proof. This follows immedi-
ately from the last subsection. It
is a case of what is known as
Pascal’s theorem.

Z 1 W 2

Z 3

W 1

Z 2

W 3

Figure 10.15. A case of Pascal’s theorem

COROLLARY.If(Z 1 ,W 1 ),(Z 2 ,W 2 ),(Z 3 ,W 3 ),(Z 4 ,W 4 )are four distinct pairs of

points all on a circle and such that

Z 1 W 2 ‖W 1 Z 2 , Z 2 W 3 ‖W 2 Z 3 , Z 3 W 4 ‖W 3 Z 4 ,

then Z 1 W 4 ‖W 1 Z 4.