Geometry with Trigonometry

112 Circles; their basic properties Ch. 7

T 2 ∈[T 1 ,U. Similarly every point ofCis also in the closed half-plane with edge

PT 2 and which containsT 1. It follows that every point ofClies in the interior region

IR(|T 1 PT 2 ). Now every point of[P,Ris in this interior region and soQis. It follows

thatQ∈[T 1 ,T 2 ]and so by 7.1.1 is interior to the circle; we thus must haveQ∈[R,S]

by 7.1.1 again.

We letx=|P,R|,y=|P,S|,z=|P,Q|, and without loss of generality assume

|P,R|<|P,S|so thatx<y.AsPis outside the circle it is outside the segment[R,S];

asQis on the segment[R,S], it follows that 0<x<z<y. Then in turn

xy=

1

2

(x+y)z,

2

z

=

1

x

+

1

y

,

1

x

−

1

z

=

1

z

−

1

y

,

z−x

xz

=

y−z

zy

,

x

z−x

=

y

y−z

,

|P,R|

|R,Q|

=

|P,S|

|S,Q|

In the above we have assumed thatlis not the lineOP.Whenitiswehavea

simple case;lcuts the circle in pointsR 1 ,S 1 such that[R 1 ,S 1 ]is a diameter. Then

taking|P,R 1 |<|P,S 1 |, with the notation of 7.4.1 we have that

|S 1 ,P|

|P,R 1 |

=

b+a

b−a

,

|S 1 ,U|

|U,R 1 |

=

a+a^2 /b

a−a^2 /b

,

and these are equal.

7.5 Anglesstandingonarcsofcircles...................

7.5.1 .....................................

Let P,Q,R,S be points of a circleC(O;k)such that R and S are on one side of the

line PQ. Then|∠PRQ|◦=|∠PSQ|◦.

(i)When O∈PQ,|∠PRQ|◦=90.

(ii)When O∈PQ and R is on the same side of PQ as O is, then|∠PRQ|◦=

1

2 |∠POQ|

◦.

(iii)When O∈PQ and R is on the opposite side of PQ from O, then|∠PRQ|◦is
equal to half of the degree-measure of the reflex-angle with support|POQ.
Proof. We are given thatR∈PQ,S∈PQ.NowS∈QRas by 7.1.1 a line cannot
meet the circle in more than two points; for this reason alsoS∈RP. ThusScannot be
on a side-line of the triangle[P,Q,R]. Moreover, neither canSbe in[P,Q,R]but not
on a side, as then by the cross-bar theorem we would haveS∈[P,V]for some point
Vin[Q,R]but not at an end-point. ThenVwould be interior to the circle andPis on
the circle, so by 7.1.1 every point of[P,V], other thanP, is interior to the circle; this
would makeSinterior to the circle whereas it is on it.
Thus asS∈[P,Q,R]we must have at least one of the following :-
(a)Sis on the opposite side ofQRfromP,
(b)Sis on the opposite side ofRPfromQ,
(c)Sis on the opposite side ofPQfromR,