Euclid's Elements of geometry, the first three books (the fourth, fifth, and sixth books) tr. from the Lat. To which is added, A compendium of algebra (A compendium of trigonometry).

a part to the whole, which is absurd; therefore A is not the centre of both circles, and it can similarly be proved, that no other point is the centre of both circles.

PROPOSITION VI. THEOREM.

If two circles (ABC, ABF) touch, one being within the other, they will not have the same centre.

For, if it be possible, let A be the centre of both circles, and draw AB to the point of contact, and also draw ACF where they do

not cut.

Therefore, because A is the centre of the circle ABC, AB shall be equal to AC (by Def. 15. B. 1); and because A is the centre of the circle ABF, AB will be equal to AF (by Def. 15. B. 1); therefore AC is equal to AF (by Ax. 1), a part to the whole, which is absurd: therefore A is not the centre of both circles, and it may be similarly shown, that no other point is the centre of both circles.

1st. If a point be taken within a circle, which is not the centre, and from it right lines be drawn to the circumference, the greatest will be that which passes through the centre.

2ndly. The remaining part of the diameter will be the least.

3rdly. Those lines which make equal angles with the diameter are equal.

4thly. That line which is nearer to the diameter is greater than those more remote.

5thly. More than two lines cannot be drawn equal.

PART 1.-CB passing through the centre, is greater

than any other line CD. For draw AD, and AB is equal to AD (by Def. 15. B. D 1), therefore if the common line CA be added to both, CA and AD together will be equal to CB, but CA and AD E together are greater than CD (by Prop. 20. B. 1), and therefore CB is greater

than CD.

PART 2.—The remaining part of the diameter CF, is less than any other line CE. For draw AE, and AC and CE together are greater than AE (by Prop. 20. B. 1), and therefore greater than AF; take away from both the common part AC, and CE will be greater than CF.

PART 3.-The right lines CL and CD which make equal angles with CB are equal. But if not, if it be possible, let either of them CL be greater, and make CG equal to CD, and draw AD and AG. Therefore in the triangles ACG and ACD, the side AC is common, and CG is equal to CD (by Constr.), and the angle ACG is equal to the angle ACD (by Hypoth.), and therefore the sides AG and AD are equal (by Prop. 4. B. 1), but the right line AD is equal to AO, and therefore AG is equal to AO, a part equal to the whole, which is absurd. Therefore neither CL nor CD is greater than the other, and therefore they are equal.

PART 4.-CD or CL which is nearer the diameter is greater than any other line CE which is more remote. If the given lines CD and CE be at the same side of CB, draw AD and AE, and in the triangles CAD, CAE, the sides CA and AD are equal to the sides CA and AE, and the angle CAD is greater than the angle CAE, and therefore the remaining side CD is greater than the remaining side CE (by Prop. 24. B. 1). But if the given lines CL and CE be at different sides of CB, construct the angle ACD equal to ACL, and CD will be equal to CL (by 3 Part), but CD is greater than CE, and therefore CL is greater than CE.

PART 5.-More than two equal lines cannot be drawn. For let any three right lines be drawn from the point C to the circumference, and any one of them may be part of the diameter, and therefore greater or less than either of the remainders (by Part first and second), or two of them will be at the same side of the diameter, and therefore unequal (by Part 4.)

1st. If there be any point assumed without a circle, and right lines be drawn from it to the circumference, those making equal angles with the line passing through the centre, will be equal.

2ndly. And of those incident on the concave circumference, the greatest is that which passes through the centre.

3rdly. Of the remaining lines, that which is nearer to the line passing through the centre, is greater than the line which is more remote from the line passing through the centre.

4thly. But of those incident on the convex circumference, that line is the least, which would pass through the centre if produced.

5thly. Of the remaining lines, that, which is nearer to the least, is less than the line which is more remote from the least.

6thly. Only two equal lines can be drawn, either to the concave or convex circumference.

PART 1.—AB and AX which make equal angles with AZ, are equal.

But if not, if it be possible, let either of them AB be greater than the other, and make AE equal to AX, x and draw ZE and ZX, therefore in the triangles ZAE, ZAX the side Y ZA is common to both, and AE is

equal to AX (by Constr.), the angle ZAE is also equal to ZAX (by Hypoth.), and therefore the sides ZE and

ZX are equal (by Prop. 4. B. 1), but the right line ZO is equal to ZX (by Def. 15 B. 1), and therefore ZE is equal to ZO; a part equal to the whole, which is absurd.

Therefore neither AB nor AX is greater than the other, and therefore they are equal.

PART 2.-Of those incident on the concave circumference, that line AY which passes through the centre is greater than any other AX. Draw ZX, and ZY will be equal to ZX (by Def. 15. B. 1), and therefore if the common line AZ be added to both, AY will be equal to AZ and ZX taken together, but AZ and ZX together are greater than AX (by Prop. 20. B. 1), and therefore AY is greater than AX.

PART 3.-AB or AX which is nearer the greatest, is greater than AD which is more remote.

If the given lines AX and AD be at the same side of AY, draw ZX and ZD, and in the triangles AZX, AZD, the sides AZ, ZX are equal to the sides AZ, ZD, and the angle AZX is greater than AZD, and therefore the remaining side AX is greater than the remaining side AD (by Prop. 24. B. 1). But if the given lines AB and AD be at different sides of AY, construct the angle ZAX equal to ZAB, and AX will be equal to AB (by Part first), but AX is greater than AD, and therefore AB is greater than AD.

PART. 4. Of those falling (or incident) on the convex circumference, that line AF, which if produced would pass through the centre, is less than any other line AX. For draw ZF and ZX, and ZX and XA are greater than ZA (by Prop. 20. B. 1), and therefore if the equals ZX and ZF be taken away, AX will be greater than AF.

PART 5.-AB or AX which is nearer the least line, is less than AC which is more remote.

If the given lines AX and AC be at the same side of AZ, draw ZX and ZC; and ZC and CA taken together are greater than ZX and XA together (by Prop. 21. B. 1), therefore take away the equals ZC and ZX, and AC will be greater than AX.

But if the given lines AB and AC be at different sides of AZ, construct the angle ZAX equal to the angle ZAB, and AX will be equal to AB (by Part first), but AC is greater than AX, and therefore greater than AB.

PART 6.-Only two equal right lines can be drawn either to the concave or convex circumference.

For if any three lines be drawn, either one of them shall pass through the centre, and therefore be either greater or less than the others (by Parts 2 and 4), or two of them will be at the same side of the line passing through the centre, and therefore unequal (by Parts 3 and 5).

SCHOL.-Hence it appears easily, that any right line drawn to the concave circumference, is greater than a right line drawn to the convex; it follows that, if any three right lines are drawn from a point without a circle to its circumference, only two of them can be equal.

COR.-Hence, and from Part 5, Prop. 7, it appears, that there is no other point, except its centre, from which three equal right lines can be drawn to the circumference.

If any point be assumed within a circle, from which to the circumference more than two equal right lines can be drawn, that point shall be the centre of the circle.

For it is not different from the centre, as if it were, only two right lines could be drawn from it to the circumference (by Prop. 7. B. 3.)

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Euclid's Elements of geometry, the first three books (the fourth, fifth, and ...