PROP. I.-PROBLEM.

To find the centre of a given circle (ADB).

Sol. Take any two points A, B in the circumference. Join AB. Bisect it in C. Erect CD at right angles to AB. Produce DC to meet the circle again in E. Bisect DE in F. Then F is the centre.

F

G

base GA equal to GB, because they are drawn from G, which is, by hypothesis, the centre, to the circumference. Hence [I. VIII.] the angle ACG is equal to the adjacent angle BCG, and therefore [I. Def. XIII.] each is a right angle; but the angle ACD is right (const.); therefore ACD is equal to ACG-a part equal to the whole-which is absurd. Hence no point can be the centre which is not in the line DE. Therefore F, the middle point of DE, must be the centre.

The foregoing proof may be abridged as follows:Because ED bisects AB at right angles, every point equally distant from the points A, B must lie in ED [I. x. Ex. 2]; but the centre is equally distant from A and B; hence the centre must be in ED; and since it must be equally distant from E and D, it must be the middle point of DE.

Cor. 1.-The line which bisects any chord of a circle perpendicularly passes through the centre of the circle. Cor. 2.-The locus of the centres of the circles which pass through two fixed points is the line bisecting at right angles that connecting the two points.

Cor. 3.-If A, B, C be three points in the circumference of a circle, the lines bisecting perpendicularly the chords AB. BC intersect in the centre.

PROP. II.-THEOREM.

If any two points (A, B) be taken in the circumference of a circle-1. The segment (AB) of the indefinite line through these points which lies between them falls within the circle. 2. The remaining parts of the line are without the circle.

Dem.-1. Let C be the centre. Take any point D in AB. Join CA, CD, CB. Now the angle ADC is [I. XVI.] greater than ABC; but the angle ABC is equal to CAB [I. v.], because the triangle CAB is isosceles; therefore the angle ADC is A greater than CAD. Hence

D

E

AC is greater than CD [I. XIX.]; therefore CD is less. than the radius of the circle, consequently the point D must be within the circle (note on Î. Def. xxIII.).

In the same manner every other point between A and B lies within the circle.

2. Take any point E in AB produced either way. Join CE. Then the angle ABC is greater than AEC [I. XVI.]; therefore CAB is greater than AEC. Hence ČE is greater than CA, and the point E is without the circle.

We have added the second part of this Proposition. The indirect proof given of the first part in several editions of Euclid is very inelegant; it is one of those absurd things which give many students a dislike to Geometry.

Cor. 1.-Three collinear points cannot be concyclic.

Cor. 2.-A line cannot meet a circle in more than two points.

Cor. 3.-The circumference of a circle is everywhere concave towards the centre.

PROP. III.-THEOREM.

If a line (AB) passing through the centre of a circle bisect a chord (CD), which does not pass through the centre, it cuts it at right angles. 2. If it cuts it at right angles, it bisects it.

A

Dem.- -1. Let O be the centre of the circle. Join OC, OD. Then the triangles CEO, DEO have CE equal to ED (hyp.), EO common, and OC equal to OD, because they are radii of the circle; hence [I. VIII.] the angle CEO is equal to DEO, and they are adjacent angles. Therefore [I. Def. XIII.] each is a right angle. Hence AB cuts CD at right angles.

D

E

B

2. The same construction being made: because OC is equal to OD, the angle OCD is equal to ODC [I. v.], and CEO is equal to DEO (hyp.), because each is right. Therefore the triangles CEO, DEO have two angles in one respectively equal to two angles in the other, and the side EO common. Hence [I. xxvI.] the side CE is equal to ED. Therefore CD is bisected in E. 2. May be proved as follows:

but

OC2 = OE2 + EC2 [I. XLVII.], and OD2 = OE2 + ED2;

Hence

Observation.-The three theorems, namely, Cor. 1., Prop. I., and Parts 1, 2, of Prop. III., are so related, that any one being proved directly, the other two follow by the Rule of Identity.

Cor. 1.-The line which bisects perpendicularly one of two parallel chords of a circle bisects the other perpendicularly.

Cor. 2.-The locus of the middle points of a system of parallel chords of a circle is the diameter of the circle perpendicular to them all.

Cor. 3.-If a line intersect two concentric circles, its intercepts between the circles are equal.

Cor. 4.-The line joining the centres of two intersecting circles bisects their common chord perpendicularly.

Exercises.

1. If a chord of a circle subtend a right angle at a given point, the locus of its middle point is a circle.

2. Every circle passing through a given point, and having its centre on a given line, passes through another given point.

3. Draw a chord in a given circle which shall subtend a right angle at a given point, and be parallel to a given line.

PROP. IV.-THEOREM.

Two chords of a circle (AB, CD) which are not both diameters cannot bisect each other, though either may bisect the other.

Dem.-Let O be the centre. Let AB, CD intersect in E; then since AB, CD are not both diameters, join OE. If possible let AE be equal to EB, and CE equal to ED. Now, since OE passing through the centre bisects AB, which does not pass through the centre, it is at right angles to A it; therefore the angle AEO is right. In like manner the angle

E

B

CEO is right. Hence AEO is equal to CEO-that is, a part equal to the whole-which is absurd. Therefore AB and CD do not bisect each other.

Cor. If two chords of a circle bisect each other, they are both diameters.

PROP. V.-THEOREM.

If two circles (ABC, ABD) cut one another in any point (A), they are not concentric.

Dem.-If possible let them have a common centre at O. Join OA, and draw any other line OD, cutting the circles in C and D respectively. Then because O is the centre of the circle ABC, OA is equal to OC. Again, because O is the centre of the circle ABD, OA is equal to OD. Hence OC is equal to OD-a part equal

to the whole-which is absurd. Therefore the circles are not concentric.

Exercises.

1. If two non-concentric circles intersect in one point, they must intersect in another point. For, let O, O' be the centres, A the point of intersection; from A let fall the AC on the line 00'. Produce AC to B, making BC = CA: then B is another point of intersection.

2. Two circles cannot have three points in common without wholly coinciding.

PROP. VI.-THEOREM.

If one circle (ABC) touch another circle (ADE) internally in any point (A), it is not concentric with it.

Dem. If possible let the circles be concentric, and let O be the centre of each. Join

OA, and draw any other line OD, cutting the circles in the points B, D respectively. Then because O is the centre of each circle (hyp.), OB and OD are each equal to OA; therefore OB is equal to OD, which is impossible. Hence the circles

cannot have the same centre.

I

B