Proposition 7. Theorem.

207. In the same circle, or in equal circles, of two unequal chords, the less is at the greater distance from the centre.

Hyp. In the

ABD, let the chord

CD the chord AB, and let the s
OE, OF be drawn from the centre O

Equal chords are equally distant from the centre (206).

Because arc AG <arc AB, the pt. G must fall within

the arc AB.

.. OH will cut the chord AB at some pt. K.

Now,

And

The

OH > OK.

The whole is any of its parts (Ax. 8).

OK > OF.

is the shortest distance from a pt. to a line (58). .. a fortiori, OH > OF.

... OE > OF.

Q.E. D

208. COR. Conversely, of two chords unequally distant from the centre, the one which is at the greater distance is the less.

EXERCISE.

If two chords of a circle cut each other, and make equal angles with the straight line which joins their point of intersection to the centre, prove that the chords are equal..

Proposition 8. Theorem.

209. A straight line perpendicular to a radius at its extremity is a tangent to the circle.

B

A

Hyp. Let O be the centre of a O, OA the radius, and BC a line

to OA at A.

To prove BC tangent to the O.

Proof. In BC take any pt. D, other than A; join OD. Then, since OA is to BC,

and

.. OA < OD.

The is the shortest distance from a pt. to a line (58).

.. the pt. D is without the circle.

(Hyp.)

[192. (1)]

(182)

.. BC has every pt. except A without the O,
... BC is a tangent to the O at A.

Q. E.D.

210. COR. 1. Conversely, a tangent to a circle at any point is perpendicular to the radius drawn to that point. 211. COR. 2. The perpendicular to a tangent at the point of tangency passes though the centre of the circle.

212. COR. 3. The straight line drawn from the centre perpendicular to the tangent meets it in the point of con

tact.

213. COR. 4. Only one tangent can be drawn to a circle at a given point on the circumference.

Proposition 9. Theorem.

214. Two parallel lines intercept equal arcs on the circumference.

There may

be three cases.

E

.*. arc AE = are BE, and arc CE = arc DE, (201) ... arc AC = arc BD.

(Ax. 3) CASE II. When AB is a tangent and CD is a secant. To prove arc CE = arc DE.

Proof. Draw the radius OE to the pt.

A E

215. COR. 1. Conversely, if the arcs intercepted by two secants are equal, the secants are parallel.

216. COR. 2. The straight line joining the points of contact of two parallel tangents is a diameter,

Proposition 10. Theorem.

217. Through three given points not in the same straight line, one circumference, and only one, can be drawn.

G.

B

Hyp. Let A, B, C be the three given pts. not in a st. line.

To prove that one Oce, and only one, can be drawn through A, B, C.

Proof. Join AB, BC.

Bisect AB, BC by the Ls DF, EG.

Since AB, BC are not in the same st. line,

.. the 1s DF, EG must meet at some pt. O.

Because O is in the DF,

.. it is equidistant from A and B.

And because O is in the EG,

... it is equidistant from B and C.

Then, because O is equidistant from A, B, C,

(Пур.)

(76)

(66)

(66)

.. the Oce described with centre O and radius OA will

pass through A, B, C.

Again, only one Oce can be so described.

For if any Oce pass through A, B, C, its centre will be at once in the bisectors DF, EG, and .. at their pt. of intersection. But two st. lines cannot intersect in more

than one pt.

.. there is only one Oce that can pass through A, B, and C.

Q.E. D.

218. COR. 1. Two circumferences cannot intersect in more than two points.

219. COR. 2. Two circumferences which have three points common coincide.

RELATIVE POSITION OF TWO CIRCLES.

Proposition 11. Theorem.

220. If two circumferences intersect each other, the right line joining their centres bisects their common chord at right angles.

Hyp. Let O, O' be the centres of two Oces which intersect each other; and A, B their pts. of intersection.

To prove that the line 00' bisects AB at rt. s.

A

B

Proof. Because O and O' are each equally distant from

A and B,

.. the line 00' bisects AB at rt s.

(179)

(67)

221. COR. 1. Conversely, the perpendicular bisector of a common chord passes through the centres of both circles. 222. COR. 2. If we suppose the circles to be moved so that the point A approaches the line 00′, the pt. B will also approach the line; and since the line 00' is always perpendicular to the mid

dle of AB (220), the two points

A and B will ultimately come together on the line 00', and be united in a single point common to the two circles. The common chord AB will then be a common tangent to the two circumferences at their point of contact.

Hence, when two circumferences are tangent to each other, their point of contact is in the straight line joining their centres; and the perpendicular at this point is a common tangent to the two circumferences.