Draw the right lines F B, F D, B D.

Then [by 22. 3.] the angles C, F are equal to two right angles. But [byfuppofition] the angles C and A are also equal to two right angles. Therefore the angles A, F will be equal to one another: Which cannot be, fince [by 21. 1.] the angle F is greater than A.

Therefore, &c. Which was to be demonftrated.

PROP. IX. THE OR.

If there be any right lined figure having an even number of fides in a circle, the fum of all the angles of the figure beginning at any one, fucceeding one another according to the odd numbers, will be equal to the fum of all the angles, fucceeding one another according to the even numbers.

Let ABCDEF [fig. 1.] be a figure in a circle having fix fides: I fay,the fum of its three angles, A, C, E will be equal to the fum of its three angles B, D, F.

For join the firft angle A, and the fourth D, by a right line.

Then because [by 22. 3.] the oppofite angles ADC,

B

A

F

Fig. 1.

E

C

ABC of the quadrilateral figure ABCD are equal to the oppofites angles BAD, B C D; alfo the oppofites angles A DE, AF E of the quadrilateral figure ADEF equal to the opposite angles F A D, D FED. Therefore the fums of the angles ADC, ADE, ABC, AFE will be equal to the fum of the angles BAD, FAD, BCD FED. But the angles A DC,

A D ́E are equal to the angle CD E, and the angles B A D, FAD equal to the angle B A F. Therefore the fum of the three angles B A F, B C D, F E D will be equal to the sum of the three angles A B C, CD E, A FE.

Again, Let the figure ABCDEFGH [fig. 2.] in the circle have eight fides: I fay, its four angles A, C, E, G will be equal to its four angles B, D, F, H.

For

For divide the figure into the figure A B C D having four fides, and the figure AHGFED having fix fides, by drawing the right line A D from the first angle A to the fourth D.

B

Fig. 2.

C

Then [by 22. 3.] the angles B A D, C will be equal to the angles CDA, B, and the angles H A D, E, G [by what has been already demonftrated] equal to the angles A DE, F, H. Therefore [by adding] the angles BAD, HAD, C, E, G will be equal to the angles CDA, ADE, B, F, H: that is, fince BAD, HAD are equal to A, and ADE, CDA equal to D; A the angles A, C, E, G will be equal to the angles B, D, F, H. Again, Let the figure H ABCDEFGHIK [fig. 3.] have ten fides: I fay, its five angles A, C, E, G, I will be equal to its five angles B, D, F, H, K.

G

F

D

E

For divide the figure into two other figures having fix fides, by the right line A F, drawn from the first angle A to the fixth angle F.

E

Then [by the laft cafe but one] the angles BA F, C, F

will be equal to the angles

B, D, E FA; and the angles KAF, G, I equal to the angles AFG, H, K. Therefore [by adding] the angles BAF, KAF, C, E, G, I will be equal to the angles B, D, E FA, AFG, H, K; that is because the angles BAF, KAF are equal to the angle A, and the angles EFA, AFG equal to the

A

K

B

I

Fig. 3.

C

D

E

F

G

angle F, the five angles A, C, E, G, I will be equal to the five angles B, D, F, H, K.

After the fame manner the theorem is demonftrated, when the number of even fides is twelve, fourteen, &c. Therefore, &c. Which was to be demonftrated.

SCHO

SCHOLIUM.

Hence it appears, that the xxii. propofition of Euclid's third book is only a particular and the moft fimple cafe of this generals propofition; being that where the right-lined figure in the circle has but four fides; alfo it is easy to demonftrate that a circle paffing thro' any three angles of a right-lined figure of an even number of fides, whofe angles have the conditions as mentioned in this our propofition, will pass thro' the other angles of the right-lined figure.

PROP. X. THEO R.

In a circle if there be three tangents meeting one another, and from the point of contact of the middle tangent be drawn two right lines to the points of contact of the other two tangents, the angles. contained under the tangents taken together will be double to the angle contained under the right lines joining the point of contact.

Let the three right lines A B, BC, CD touch the circle AED in the points A, E, D, and let the middle tangent BC meet the other two in the points B, C: I fay, the angles A BE, E CD, taken together, will be double to the angle A E D.

For [by 1. 3.] find the femidiameters F A,

B

A

F

E

C

the centre of the circle, and draw F E, F D to the points of contact.

Then because [by 5. 1.] the angles FAE, FEA are equal to one another, as alfo FDE, FED. Therefore [by 32. 1.] twice the angle D FAE, together with the angle AFE, will be equal to two right angles. In like manner twice the angle FED, together with the angle EFD will be equal to two right angles. And fo [by adding] twice the angle FA E, twice the angle FED, together with the angles AFE, EFD, will be equal to four right angles. But because the femidiamcters FA, FE, FD [by 18. 3.] are at right angles to the tangents A B, B C, D C, and fince all the angles of any quadrilateral figure are equal to four right angles; therefore

the

the angles A FE, A BE of the quadrilateral figure A BEF, will be equal to two right angles; in like manner the angles DFE, DCE of the quadrilateral figure FECD will be equal to two right angles: Wherefore [by adding] the angles AFE, ABE, DFE, DCE will be equal to four right angles. Therefore twice the angle F A E, twice the angle FED, together with the angles AF E, DE F, will be equal to the angles AEF, ABE, DFE, DCE; and taking away the angles A FE, DFE, which are common, there will remain twice the angle FAE, together with twice the angle FED, equal to the angles ABE, DCE, that is, fince the angle AED is equal to the angles F A E, F E D, twice the angle A ED will be equal to the angles ABE, DCE. Therefore, &c. Which was to be demonstrated.

Corollary. Hence if each fide of a right lined figure touches a circle, that is, if a right lined figure circumfcribes a circle, and right lines be orderly drawn joining the points of contact forming a right lined figure of the fame number of fides within the circle, any two adjacent angles of the circumfcribing figure will be double to the angle of the figure within the circle, which is between them.

PROP. XI.

If each fide of any right lined figure, whofe fides are even in number, touches a circle, the fum of the first, third, fifth, &c. beginning at any one fide, and orderly proceeding, according to the odd numbers, will be equal to the fum of the fecond, fourth, fixth, &c. fides, orderly proceeding according to the even numbers.

First, let the right lined figure [fig 1.] have four fides A B, BC, AD, DC, and let thefe fides touch the circle EFGH, in the points E, F, G, H; I fay the fum of the firft fide A B, (making A B the firft, and Bc the second) and the third fide c will be equal to the fum of the fecond fide BC, and the fourth fide A D.

For fince A E touches the circle in E, the rectangle under any right line drawn from a to the convex part of the circle, and its external part will [by 37. 3.] be equal to the fquare of the tangent AE: So alfo becaufe AH touches the circle in H, the rectangle under any other

M

B

Fig. 1.

right line drawn from A, to the convexity of the circle and its external part, will be equal to the fquare of the tangent A H; wherefore the fquares of the tangents A E, AH will be equal to one another; and fo their fides A E, A H will be equal to one another. In like manner it is proved that EB will be equal to BF; FC equal to CG; and GD equal to DH: Therefore AE, BF will be equal to A B; EB, FC equal to BC; FC, GD equal to CD; and DG, A H equal to AD. Wherefore [by adding] AE, BF, FC, GD will be equal to A B, CD. But fince BC, AD are equal to EB, FC, GD,

A

E

C

F

H

G

AH, that is (because EB has been proved to be equal to BF, and H D equal to DG) equal to A E, B F, FC, GD; therefore will A B, C D be equal to B C, A D.

Again, Let the figure [fig. 2.] have fix fides, AB the firft, BC the fecond, CD, DI, IL, AL touching the circle in the points E, F, G, H, K, M: I fay AB, CD, IL will be equal to BC, DI, AL.

For from what has been already proved above, it follows that the several tangents A E, AM; BE, BF; CF, CG; DG, DH; HI, IK; KL, LM; will be each two equal to

B

E

Fig. 2.

F

G

one another. Therefore AB will be equa to AE, BF; BC equal to BE, CF; CD equal to CF, GD; DI equal to GD, HI; DIL equal to HI, LK; and AL equal to LK, AM: Therefore [by adding] AB, CD, IL will be equal to AE, BF, CF, GD, HI, LK; and B C, DI, AL equal to BE, CF, GD, HI, LK, AM ;

H

A

I

and [adding equals to equals] A B, CD,