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diameter, having bisected it in y, cut off d i equal to e h, and
D DK equal to he: So that the whole ik be equal to the whole EF; and thro' , k [by 11. 1.) draw L M, N o perpendicular to Ac; and join IN: I say I n is applied equal to e F, and parallel to G: For since LM, NO are equally distant from the centre; they will [by 14. 3.) be equal to one nother : and because they are divided in half at 1, K [by 3. 3.) for they are cut at right angles by the right line a c passing thro' the centre d, their halves L 1, KN will be equal. But because [by 28. 1.] LI, NK are also parallel ; also LN, IK [by 31. 1.] will be equal and parallel : Wherefore since ik is equal to E F and parallel to G; L N will be equal to E F, and [by 30. 1.) parallel to G: By the same reason if mo be drawn, it will also be equal to E F, and parallel to g. Which was to be demonstrated.
PROP. II. PROBL. To infcribe a triangle in a given circle, equiangular
to a given triangle b. Let the given circle be ABC, and the given triangle be DEF; it is required to inscribe a triangle in the circle A B C equiangular to the triangle de F.
Draw the right line G A H touching the circle A B C in the point A, and at the right line Ah, and at the point A, in it [by 23. 1.) make the angle HAC equal to the angle DEF: Allo at the right line GA, and at the point A in it make the angle G A B equal the angle F D E, and join
Then because the right line HAG touches the circle
А ABC, and AC is drawn
D) from the point of contact ; the angle HAC [by 32. 3.] will be equal to the angle E F A B C, which itands in the
BY alternate segment of the circle: But the angle H A c is equal to the angle DEF; and therefore the angle A B C is equal to the angle DEF. By the same reason, the angle A CB is equal to the angle FDE : Therefore the remaining angle B A C will be [by 32. 1.) equal to the remaining angle EFD; and so the triangle ABC is equiangular to the triangle D E F, and [by 3. def. 4.] is inscribed in the circle A B Ca
Wherefore a triangle is inscribed in a circle, equiangular to a given triangle. Which was to be done.
. This problem may be otherwise constructed thus: Let ABC be the given circle, whose centre is H, and d e f a given triangle to which a triangle is to be inscribed equiangular to the given triangle de F: Draw any semidiameter 1 A, and at the centre h with the semidiameter' n A make [by 23. 1.) the angle A H B double to one of the angles E D F of the given triangle. Apply the right line AB. At the point A with the
right line A B make the angle Bac equal
to the angle ppp of E
the given triangle. H
Apply the right line D
BC. Then will the triangle AB be inscribed in the given circle A B C equiangular to the given
triangle De F. For fince [by construction) the angle bh a at the centre is double to the angle EDF; and the angle BCA at the circumference [by 20. 3.) is one half that angle BHA; therefore the angle BCA will be equal to the angle ed .
But the angle BAC (by construction) is equal to the angle pie. Therefore [by 32. 1.) the remaining angle A B C will be equal to the remaining angle der of the given triangle. Wherefore the triangle ABC is equiangular to the triangle D E F, and is inscribed in the circle A 'BC.
Hence an equilateral triangle may be B
easily inscribed in a given circle ; for it C is but applying A D equal to the semidi
ameter A H of the circle ; and then again H
applying BD equal to Ah, and joining the
right line AB, which will be one side of D
the equilateral triangle ABC inscribed in
the circle. The demonstration is easy, А
from what has been said above, and
from prop. xxvii, xxviii. lib. iii.
PROP. III. PROBL. To circumscribe a triangle about a given circle equi,
angular to a given triangle. Let ABC be the given circle, and DEF the given tri
angle; it is required to circumscribe a triangle about the given circle ABC, equiangular to the given triangle D EF.
Produce E F [by 2. poft.] both ways to the points H and G, and [by 1. 3.] find the centre k of the circle A B C, and any how drawing the
D right line KB, make [by 23. 1. ] at the line K e, and at the point K in it, the angle B K A equal to the an- А.
G E I H gle D EG, and the angle BK c equal to the angle DFH; and draw [by 17. M
B N 3.) the right lines 1 AM, MBN, NCL touching the circle A B C in the points A, B, C.
Then because the right lines L A M, MBN, NCL touch the circle A B C in the points A, B, C, and the right lines KA, KB, KC are drawn from the centre k to the points A, B, C, the angles at the points A, B, C, will [by 18. 3.] be right angles ; and because the four angles of the quadrilateral figure A M B K, are [by 32. 1.] equal to four right angles (for the quadrilateral figure A MKB may be divided into two triangles) whereof the angles KAM, KBM are each one right angle; the remaining angles A K B, A M B will be equal to two right angles. But [by 13. 1.) the angles DEG, DEF are also equal to two right angles : Wherefore the angles AKB, AMB, are equal to the angles deg, DEF. But A K B is equal to DEG: Therefore the remaining angle A MB, will be equal to the remaining angle DEF. After the same manner we demonstrate, that the angle L N M is equal to the angle D FE; and so the remaining angle ML N [by 32.1.] is equal to the remaining angle e DF: Therefore the triangle L M N is equiangular to the triangle D E F, and (by 4. def. 4.) it is circumfcribed about the circle AB C.
Therefore a triangle is circumscribed about a given circle equiangular to a given triangle. Which was to be done.
PROP. IV. PROBL. To inscribe a circle in a given triangle. Let the given triangle be ABC : it is required to inscribe a circle in the triangle A BC.
Divide [by 9. 1.] the angles A B C, B C A each into two equal parts, by the right lines BD, CD, meeting one another in the point D, and from D draw [by 12. 1.] the perpendiculars D E, DF, D G to the sides A B, BC, CA, of the triangle.
Then because the angle A B D is equal to the angle CBD, for the angle A B C is bisected, and the right angle BED is also equal to the right angle B F D, there are two triangles EBD, DBF А having two angles of the one equal to two angles of the other, and one side of the one equal to one side of the other, viz. the side B D opposite to one of the equal angles common to them both :
D Therefore [by 26. 1.] the remaining
G sides of the one will be equal to the remaining sides of the other; that is, DE will be equal to DF; also by the fame B F с reason the right line D G will be equal to DF: wherefore if a circle be described with the centre D, and either of the distances D E, DF, DG, it will pass thro' the remaining points, and touch the right lines A B, BC, CA, because the angles at E, F, G, are right angles: for if it cuts them the right line drawn from the end of a diameter at right angles to it, will fall within the circle ; which [by 16. 3.] is absurd: Therefore the circle described with the centre D and either of the distances DE, DF, DC will not cut the right lines A B, BC, CA; wherefore it will touch them, and [by 5. def. 4.] be inscribed in the triangle ABC.
Therefore the circle E F G is inscribed in the given triangle a Bc. Which was to be done.
PROP. V. PROBL.
Let ABC be a given triangle: it is required to circumfcribe a circle about the given triangle A B C:
Divide [by 10. 1.] AB, A c each into two equal parts in the points D, E, and (by 11. 1.] draw D F, EF from the points D, E at right angles to A B, AC; these will either
meet within the triangle ABC, in the side B C, or without the triangle A B C.
First let them meet within the triangle in the point F, and join B F, FC, FA: Then because a D is equal to DB, and D F is both common and at right angles, the base A F [by 4. 1.] will be equal to the base F B. In like manner we demonstrate that cr is equal to Fa, and fo BF is equal to Fc: wherefore the three right lines F A, FB, FC are equal to one another : Consequently a circle described from the centre F with either of the distances F A, FB, FC will pass thro' the remaining points; and (by 6. def. 4.] the circle will be circumscribed about the triangle ABC. Let it be circumscribed as A B C.
Secondly, Let DF, EF meet at F in the right line B C, as in the second figure; and join A F. We demonstrate after the fame manner that the point F is the centre of a circle circumscribed about the triangle A B C.
Lastly, Let DF, EF meet in the point F without the triangle A BC, as in the third figure; and join A F, FB, FC: Then because AD is equal to D B, and D F is common and at right angles, the base A F [by 4. 1.] will be equal to the base FB. Also after the fame manner we de. monstrate that c F is equal to FA; and fo B F is equal to FC; therefore again a circle described with the centre F, and either of the distances F A, TB, FC will pass thro' the