250. In the same circle or in equal circles, if two chords are unequal, they are unequally distant from the centre; and the greater chord is at the less distance.

In the circle whose centre is 0, let the chords AB and CD be unequal, and AB the greater; and let OE be perpendicular to AB and OF perpendicular to CD.

Proof. Suppose AG drawn equal to CD, and OH ¦ to AG.

.. ZOHE, the complement of AHE, is less than

Ax. 7

§ 152

OEH,

Ax. 5

§ 153

§ 249

Q. E.D.

OH OF.

=

.. OE < OF.

Ex. 91. The perpendicular bisectors of the sides of an inscribed poly gon are concurrent (pass through the same point).

PROPOSITION VIII. THEOREM.

251. CONVERSELY: In the same circle or in equal circles, if two chords are unequally distant from the centre, they are unequal; and the chord at the less distance is the greater.

In the circle whose centre is 0, let AB and CD be unequally distant from 0; and let OE, the perpendicular to AB, be less than OF, the perpendicular to CD.

Proof. Suppose AG drawn equal to CD, and OHL to AG.

252. COR. A diameter of a circle is greater than any

OHE, is greater than

Ax. 5

§ 153

=

AG.

§ 245

Ax. 6

Const.

Q. E.D.

other chord.

PROPOSITION IX. THEOREM.

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

Let MB be perpendicular to the radius OA at A.
To prove that MB is a tangent to the circle.

Proof. From O draw any other line to MB, as OH.

Then

OH > OA.

.. the point H is without the circle.

$ 97

§ 216

Hence, every point, except A, of the line MB is without the circle, and therefore MB is a tangent to the circle at A. § 220

Q. E. D.

254. COR. 1. A tangent to a circle is perpendicular to the radius drawn to the point of contact.

For OA is the shortest line from 0 to MB, and is therefore 1 to MB (§ 98); that is, MB is 1 to OA.

255. COR. 2. A perpendicular to a tangent at the point of contact passes through the centre of the circle.

For a radius is to a tangent at the point of contact, and therefore a erected at the point of contact coincides with this radius and passes through the centre.

256. COR. 3. A perpendicular from the centre of a circle to a tangent passes through the point of contact.

CASE 1. Let AB (Fig. 1) be a tangent at F parallel to CD, a secant.

CASE 2. Let AB and CD (Fig. 2) be parallel secants.

Proof. Suppose EF to CD and tangent to the circle at M.

CASE 3. Let AB and CD (Fig. 3) be parallel tangents at E and F.

Suppose GH drawn to AB.

PROPOSITION XI. THEOREM.

258. Through three points not in a straight line one circumference, and only one, can be drawn.

B

Let A, B, C be three points not in a straight line.

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

At the middle points of AB and BC suppose Is erected. These Is will intersect at some point O, since AB and BC are not in the same straight line.

The point O is in the perpendicular bisector of AB, and is therefore equidistant from A and B; the point O is also in the perpendicular bisector of BC, and is therefore equidistant from B and C.

§ 160

Therefore, is equidistant from A, B, and C; and a circumference described from O as a centre, with a radius OA, will pass through the three given points.

The centre of a circumference passing through the three points must be in both perpendiculars, and hence at their intersection. As two straight lines can intersect in only one point, O is the centre of the only circumference that can pass through the three given points.

Q. E.D.

259. COR. Two circumferences can intersect in only two points. For, if two circumferences have three points common, they coincide and form one circumference.