AG [Def. 10. 11, therefore FC and DG together are greater than FG, the part than the whole, which is absurd. Therefore the right line, which joins the centres of the circles ABC, AED, cannot pass otherwise than through the contact A, and, of course, passes through it.

PROP. XIII. THEOR.

One circle cannot touch another, either within or without, in more points than one.

Let the circles AC, AD, if possible, touch each other inwardly in two points A and B; find the centres E, F of these circles (1. 3), which are different [6. 3]; draw EF, which produce to pass through one of the contacts as A (11. 3), and draw EB and FB.

Because FA is equal to FB (Def. 10. 1), adding to each FE, AE is equal to BF, FE; but BF, FE are together greater

BA

D

than BE [20. 1], therefore AE is greater than BE; but, because E is the centre of the circle AC, AE is equal to BE (Def. 10. 1); therefore AE is both equal to, and greater than, BE, which is absurd.

But if the two points of contact A, B be at opposite parts of the right line joining the centres E, F, the right line AB is a diameter of both the circles AC and AD, and AF is equal to FB (Def. 10. 1), and therefore greater than EB, of course AE is greater than EB; but AE is also equal to EB (Def. 10. 1), which is absurd.

Lastly, let the circles AC, AD, if possible, touch each other externally in two points A, B ; draw EF joining the centres E, F, and passing through one of them A [11. 3], and join EB,

BF.

Then is EA equal to EB, and AF to BF (Def. 10, 1); therefore EF

B

E

A

E

F

D

B

one side of the triangle EBF is equal to the other two sides EB BF, which is absurd (20. 1).

In no case therefore, do two circles touch each other in more points than one.

PROP. XIV. THEOR.

Equal right lines (AB, CD), inscribed in a circle (ABDC), are equally distant from the centre; and those, which are equally distant from the centre, are equal.

Take E the centre of the circle ABDC (1. 5), join EA, EC, and draw EF, EG perpendiculars to AB, CD.

Because AB, CD are equal (Hyp.), and bisected by the perpendiculars EF, EG (3. 3), the right lines AF, CG are equal (Ax. 7); therefore their_squares are equal (Cor. 3. 34. 1); also EA, EC are equal (Def. 10. 1), and therefore their

A

طع

squares [Cor. 3. 34. 1]; but, because the angle AFE is right, the square of AE is equal to the squares of AF, FE [47. 1], and, for the like reason, the square of EC is equal to the squares of CG, GE; therefore the squares of AF, FE are equal to the squares of CG, GE; taking from each the equal squares of AF, CG, the squares of EF and EG are equal [Ax. 3], and therefore the right lines EF, EG themselves [Cor. 1. 46. 1], and, of course, AB, CD are equally distant from the centre [Def. 4. 3].

Let now AB, CD be equally distant from the centre, and, of course, EF, EG equal [Def. 4. 3], their squares are also equal (Cor. 3. 34. 1); and because EA, EC are equal, their squares are equal (Cor. 3. 34. 1); but the square of EA is equal to the squares of EF, FA, and the square of EC to the squares of EG, GC [47. 1]; therefore the squares of EF, FA are equal to the squares of EG, GC; taking away the equal squares of EF, EG, the squares of AF, CG are equal [Ax. 3], and therefore these right lines themselves [Cor. 1. 46. 1]; but, because EF, EG bisect AB, CD [3. 3], the right lines AB, CD are double of the equals AF, CG, and therefore equal to each other [Ax. 6].

PROP. XV. THEOR

The diameter (AD) is the greatest right line in a circle (AB, CD); and, of all others, that (BC) which is nearer to the centre (E), is greater than one more remote (FG); and the greater (BC) is nearer to the centre than the less (FG).

From the centre E draw EH, EK perpendiculars to BC, FG, and join EB, EC, EF; the right line EA is equal to EB, and ED to EC (Def. 10. 1), therefore AD is equal to EB and EC; but EB and EC are greater than BC (20. 1), therefore AD is greater than BC.

D

Let now BC be nearer the centre than FG, and, of course, EH less than EK (Def. 5. 3), the right line BC is greater than FG. For the squares of BH, HE are together equal to the square of EB (47. 1), or of EF, its equal (Def. 10. 1); also the squares of FK, KE are equal to the square of EF [47. 1]; therefore the squares of BH, HE are equal to the squares of FK, KE; whence, the square of EH being less than the square of FK, because EH is less than EK, the square of BH is greater than the square of FK, and, of course, BH greater than FK; whence, BC, FG being_doubles of BH, FK [3. 3], the right line BC is greater than FG.

Lastly, let BC be greater than FG, then is BC nearer to the centre than FG. For, because BC is greater than FG [Hyp.], and BH, FK are the halves of BC, FG (3. S), the right line BH is greater than FK; whence, the squares of BH, HE, and of FK, KE, being each equal to the square of the radius EB or EF [47. 1, and therefore to each other, the square of EH is less than the square of EK, and, of course, EH less than EK, and BC nearer to the centre than FG [Def. 5. 3].

PROP. XVI. THEOR.

A right line, drawn from the extremity (A), of a diameter (AB), of a circle (ABC), perpendicular to it, falls entirely without the circle, and no right line can be drawn, between that right line and the circumference, so as not to cut the circle.

Part 1.-For if the right line, drawn from A at right angles to AB, does not fall entirely without the circle, let it, if possible, meet the circumference in some other point, as C, and from the centre D draw DC.

Because, in the triangle DAC, the sides DA, DC are equal (Def. 10. 1), the

B

D

angles DAC, DCA are also equal (5. 1), but DAC is a right angle (Hyp.), therefore DCA is a right angle, and the angles DAC, DCA together equal to two right angles, which is absurd (17. 1). Therefore a right line drawn from A, at right angles to AB, does not meet the circle in any other point, and therefore falls entirely without it.

Part 3.-Let now the right line AE be that which is drawn from A, at right angles to AB, which, by part 1. of this, falls entirely without the circle; there cannot be drawn between AE and the B circumference, a right line, so as not to cut the circle.

For, if possible, let AF be such, and the angle DAE being right, DAF is

D

H

acute; from D draw DG perpendicular to AF, meeting the circumference in H; and since, in the triangle DAG, the right angle DGA is greater than the acute angle DAG, the side ĎA is greater than DG (19. 1); but DH is equal to DA (Def. 10. 1); therefore DH is greater than DG, the part than the whole, which is absurd. Therefore no right line can be drawn between AE and the circle, so as not to cut it.

Cor.-Hence it appears, that a right line, drawn from the extremity of the diameter of a circle, at right angles to it, touches the circle (Def. 1. 3), and that it touches it only in one point.

PROP. XVII. PROB.

From a given point which is not within a given circle (BCD), to draw a right line to touch it.

First, let the given point be without the circle, as A; find its centre E (1. 3), join EA, and from the centre E, at the distance EA, describe the circle AFG; through D, draw DF perpendicular to AE (11. 1), meeting the circle AFG in F, and draw EBF and AB; the right line AB touches the circle BCD.

G

A

D

F

For in the triangles EBA, EDF, the sides EB, EA are severally equal to ED, EF (Def. 10. 1), and the angle E common, therefore the angle EBA is equal to the angle EDF (4. 1); but EDF is a right angle (Constr.), therefore EBA is a right angle, and, of course, AB touches the circle BCD (Cor. 16. 3), being drawn from the given point A.

Let now the given point be in the circumference of the given circle BDC, as the point B ; draw BE to the centre E, and BA at right angles to BE; the right line BA touches the eircle (Cor. 16. 3).

PROP. XVIII. THEOR.

If a right line (DE) touch a circle (ABC); a right line (FC), drawn from the centre (F), to the contact (C), is perpendicular to the tangent.

If FC be not perpendicular to DE, from F, draw FBG perpendicular thereto (12. 1), then, because FGC is a right angle, FCG is acute (17.1), therefore the side FC is greater than FG (19. 1); but FB is equal to FC (Def. 10. 1), therefore FB is greater than FG, the part than the whole, which is absurd; therefore FG is not perpendicular to D

B

E

C G

DE. In like manner it may be shewn, that no other right line but FC, is perpendicular to it, therefore FC is perpendicular to DE.