PROP. III. PROBL.

To infcribe a trapezium in a given circle equiangular to a given trapezium, and one of whose fides fall be of a given length. But this length must be lefs than the diameter of the given circle, and two of the oppofite angles of the given trapezium must be equal to two right angles.

Let the given circle be A B C D, the given right line I K, lefs than the diameter of the circle, and the given trapezium EFGH, having two of its oppofite angles EFG, EHG equal to two right angles; it is required to infcribe a trapezium in the circle A B C D equiangular to the given trapezium E F G H having one of its fides equal to the given right line I K.

From the given circle A B C D cut off [by 34. 3.] the fegment A B C containing an angle equal to the angle E F G of the given trapezium E F G H; draw the right line AD [by 1. 4.]

equal to the

I

given right

B

line I K, and

K

F

G

line AD, and

at the given point A in it, make the an

D

E

H

gle D A B equal to the angle FEH of the given trapezium E F GH; and let the fide A B of this angle cut the circumference of the given circle in the point B. Join the points B, C, D. And the trapezium A B C D will be that required.

For because all the angles in the same segment of a circle [by 21. 3.] are equal to one another, the angle A B C will be equal to the angle EFG, fince [by conftruction] the fegment A B C of the circle contains an angle equal to the angle E F G. And because [by 22. 3.] the angles A B C, A DC are equal to two right angles, and the angles EF G, EHG of the given trapezium are alfo [by fuppofition] equal to two right angles. Therefore will the angle ADC,

be

be equal to the angle E HG of the given trapezium E F G H. But the angle BAD [by conftruction] is equal to the angle HEF of the trapezium. Wherefore the three angles ABC, ADC, BAD of the trapezium A B C D will be refpectively equal to the three angles EFG, EH G, FEH of the given trapezium E F G H. And fo because all the angles of any trapezium are [by 3. addit. to lib. 1.] equal to four right angles, the fourth angle B CD of the trapezium A B C D will be equal to the fourth angle F G H of the given trapezium EFG H. Therefore the trapezium A B C D will be equiangular to the given trapezium, having one fide A D equal to the given right line IK; and it is infcribed in the given circle A B C D.

Therefore, &c. Which was to be done.

To circumfcribe a trapezium about a given circle, equiangular to a given trapezium.

Let NOL P be the given circle, whofe centre is M, and let E FGH be a given trapezium. It is required to circumscribe a trapezium about the given circle NOLP equiangular to the given trapezium EFGH.

Draw [by 17. 3.] from an affumed point without the circle the right line QN D touching the circle in the point N. At the right line QD, and at the point Q in it, make [by 23. 1.] the angle DQR equal to the angle E of the given trapezium; and from the centre M of the circle draw [by 12. 1.] the perpendicular M R to QR, the fide of the angle meeting the circumference of the circle in the point o; thro' o draw [by 31. 1.] A O B parallel to QR. Then [by 16. 3.] the right line AB will touch the circle

in o; and because A o, QR, [by conftruction] are parallel : Therefore [by 29. 3.] the angle O AD will be equal to the angle RQD. But the angle RQD [by conftruction] i equal to the angle E of the trapezium. Therefore the angle O AN will be equal to the angle E of the trapezium. Again, by a like operation, draw the right line DP to touch the circle in P, making an angle QDC with the right line qb equal to the angle H, of the given trapezium. And by a further like operation draw B C to touch the given circle in L, making the angle A B C with A B equal to the angle F of the given trapezium, and let в C meet DC in the point c. I fay A B C D will be the trapezium required.

For it has been already proved that the angle B A D of the trapezium A B C D is equal to the angle E of the given trapezium EFGH. So alfo does it appear [by conftruction] and from what has been already proved, that the angles ABC, A DC of the trapezium A B C D are each equal to the angles F, H of the given trapezium: Therefore the remaining angle B C D of the trapezium A B C D will be [by 3. addit. to book 1.] equal to the angle G of the trapezium E F G H. Therefore the trapeziums A B C D, EFGH are equiangular. But [by conftruction] the trapezium ABCD touches the given circle O LPN in the points 0, L, P, N.

Therefore, &c. Which was to be demonftrated.

PROP. V. PROBL.

To infcribe a regular heptagon, or figure having feven equal fides, in a given circle, partly by trial.

Let D C B be a given circle, whofe centre is A. It is required to infcribe a regular heptagon in the given circle DC B, partly by trial.

Draw the diameter DB, and affume [by 15.4.] the arch B C of a regular hexagon. From c draw [by 12. 1.] the right line C E perpendicular to the femidiameter A B of the circle, and draw another femidiameter A F [by trial, or turning a ruler about the centre A, &c.] fo cutting the circumference of the circle in the point F, and the

perpendicular c E in 1, that drawing the right linei в F, the fame fhall be equal to 1 F.

Join BI.

B

This done, make the arch BL equal to the arch B F, and draw the right line FL. I fay FL will be the fide of a regular heptagon infcribed in the given circle D C B. One fide F L being thus found, the others will be obtained by carrying that fide fix times about the circumference of the given circle.

Then fince the fides A B, A F are equal to one another, the angles AFB, A BF [by 5. 1.] will be equal to one another. And fince [by cor. 15. 4.] the fide of the regular hexagon infcribed in the circle is equal to the femidiameter A C; and because C E is perpendicular to A B, [by conftr.] therefore [by 26. 1.] A E will be equal to E B; AI equal to IB, and the angle I A E equal to the angle IB E. But fince IF, FB [by conftr.] are equal to one another, the angles F I B, FB 1, [by 5. 1.] will be equal to one another. But [by 32. 1.] the outward angle FIB of the triangle AIB, to which FBI is equal, is equal to both the inward angles I A B, I BA: To the angle FBI add the angle I B A, that is, FA B. Then will the angle A B F, or A FB equal to it, be thrice the angle FAB. Again, becaufe [by 20. 3.] an angle at the centre of a circle is double to an angle at the circumference, the angle D A F will be double to the angle ABF. And fince the angle A B F is three times the angle FA B. Therefore the angle D A F will be fix times the angle FAB; and fo the angle FAB will be one seventh part of two right angles. Wherefore [by 26. 3.] the arch FB will be one seventh part of the arch of the femicircle BCD. And accordingly [by conftruction] the arch FL will be one feventh part of the whole circumference. Therefore, &c. Which was to be demonstrated. SCHOLIUM.

This problem is not to be refolved geometrically; that is, by a right line and a circle. Its folution requires a curve

line of an higher order, as a conic fection, or the conchoid. It arifing, as the algebrifts call it, to a cubic equation. For if the femidiameter of the given circle be called r, and the fide of the given regular heptagon infcribed in it be called x, one of the roots of this cubic equation x3+4rxx+3rrx—r3=0, will give the value of the fide of the regular heptagon infcribed in the circle.

PROP. VI. PROBL.

To infcribe a regular nonagon, or figure of nine equal fides, in a given circle, partly by trial.

Let FDAB be a given circle, whofe diameter is F B, and centre c. It is required to infcribe a regular nonagon in the given circle F D A B.

Draw the diameter B F of the circle, which continue out to E. Take [by 15. 4.] the arch A B equal to the fixth part of the whole circumference of the circle. From A draw the right line A D E, fo cutting the circumference of the circle in the point D, and the continuation of the diameter BF in the point E, that the part E D of the right line E A be equal to the femidiameter F C or C B of the circle. But this is to be done by an eafy and expeditious trial. When this is done from the point F, draw [by 31. 1.] the right line FG parallel to the right line È A cutting the circum

ference of the circle in the point G. Join BG. I fay BG will be one fide of a regular nonagon, or figure having nine equal fides, infcribed in the given circle F D A B ; which apply'd nine times

E

A

round the circumference of the given circle will be that required.

For join the femidiameters C A, C D.

Then fince [by conftruction] E D is equal to DC; the angle DCE [by 5. 1.] will be equal to the angle DE C.