If a straight line touch a circle, the straight line drawn from the centre to the point of contact must be perpendicular to the line touching the circle.

Let the st. line DE touch the O ABC in the pt. C.

Find the centre, and join OC.

Then must OC be 1 to DE.

For if it be not, draw OBF1 to DE, meeting the Oce in B.

.. OB is greater than OF, which is impossible;

.. OF is not to DE, and in the same way it may be shewn that no other line drawn from O, but OC, is 1 to DE;

.. OC is to DE.

Q. E. D.

Ex. If two straight lines intersect, the centres of all circles touched by both lines lie in two lines at right angles to each other.

NOTE. Prop. XVIII. might be stated thus :-All radii of a circle are normals to the circle at the points where they meet the circumference.

If a straight line touch a circle, and from the point of contact a straight line be drawn at right angles to the touching line, the centre of the circle must be in that line.

B

E

Let the st. line DE touch the O ABC at the pt. C, and from Clet CA be drawn to DE.

Then

Then must the centre of the ○ be in CA.

For if not, let F be the centre, and join FC.

DCE touches the O, and FC is drawn from centre

In the same way it may be shewn that no pt. out of CA can be the centre of the ;

.. the centre of the O lies in CA.

Q. E. D.

Ex. Two concentric circles being described, if a chord of the greater touch the less, the parts of the chord, intercepted between the two circles, are equal.

NOTE. Prop. XIX. might be stated thus :-Every normal to a circle passes through the centre.

PROPOSITION XX. THEOREM.

The angle at the centre of a circle is double of the angle at the circumference, subtended by the same arc.

Let ABC be a O, O the centre,

BC any arc, A any pt. in the Oce.

Then must / BOC = twice ▲ BAC.

First, suppose O to be in one of the lines containing the

L BAC.

Then OA =

OC,

..LOCA

= 40AC;

.. sum of 48 OCA, OAC = twice ▲ OAC.

I. A.

I. 32.

Next, suppose O to be within (fig 1), or without (fig. 2) the

L BAC.

Fig. 1.

Fig. 2.

Join A0, and produce it to meet the Oce in D.

Then, as in the first case,

.., fig. 1, sum of 48 COD, BOD = twice sum of 48 CAD,

BAD,

that is, 4 BOC = twice 4 BAC.

=

twice difference

And, fig. 2, difference of 48 COD, BOD of 48 CAD, BAD, that is, ▲ BOC = twice ▲ BAC.

Q. E. D.

Ex. 1. The centre of the circle CBED is on the circumference of ABD. If from any point A the lines ABC and AED be drawn to cut the circles, the chord BE is parallel to CD.

Ex. 2. From any point in a straight line, touching a circle, a straight line is drawn through the centre, and is terminated by the circumference; the angle between these two straight lines is bisected by a straight line, which intersects the straight line joining their extremities. Shew that the angle between the last two lines is half a right angle.

NOTE 2. On Flat and Reflex Angles.

We have already explained (Note 3, Book I., p. 28) how Euclid's definition of an angle may be extended with advantage, so as to include the conception of an angle equal to two right angles and we now proceed to shew how the Definition given in that Note may be extended, so as to embrace angles greater than two right angles.

:

Let WQ be a straight line, and QE its continuation.

Then, by the Definition, the angle made by WQ and QE, which we propose to call a FLAT ANGLE, is equal to two right angles.

Now suppose QP to be a straight line, which revolves about the fixed point Q, and which at first coincides with QE.

When QP, revolving from right to left, coincides with QW, it has described an angle equal to two right angles.

When QP has continued its revolution, so as to come into the position indicated in the diagram, it has described an angle EQP, indicated by the dotted line, greater than two right angles, and this we call a REFLEX ANGLE.

To assist the learner, we shall mark these angles with dotted lines in the diagrams.

Admitting the existence of angles, equal to and greater than two right angles, the Proposition last proved may be extended, as we now proceed to shew.