« ZurückWeiter »
a triangle that shall have its angles equal to those of the triangle ABC. r | Assuming any point D in the circumference of the circle, draw (III, 22.) the tangent EDF, and make the angles EDG, FDH equal to BCA, BAC, and join GH: The triangle GDH is equiangular to ABC. For EF being a tangent, and DG drawn from the point of contact, the angle EDG, which was made equal to BCA, is equal to the angle DHG in the alternate segment (III. 21.); consequently DHG is equal to BCA. And for the same reason, the angle DGH is equal to BAC; wherefore (I. 30.) the remaining angle GDH of the triangle GHD is equal to the remaining angle ABC of the triangle ACB, and these triangles are mutually equian
PROP. XII. PROB.
About a given circle, to describe a triangle equiangular to a given triangle.
Let GIH be a circle, about which it is required to describe a triangle, having its angles equal to those of the triangle ABC.
Draw any radius FG, and with it make (I. 4.) the angles GFI, GFH equal to the exterior angles BAE, BCD of the triangle ABC, and, from the points G, I, and H
draw the tangents KM, KL, and LM to form the triangle KLM : This triangle is equiangular to ABC. For all the angles.of the quadrilateral figure KIFG being equal to four right angles, and the angles KIF and KGF being each a right angle (III. 20.), the remaining angles GKI and GFI are together equal to two right angles, and consequently equal to the angles BAC and BAE on the same side of the straight line ED. But the angle GFI was made equal to BAE; whence GKP is equal to CAB. In like manner, it may be proved that the angle GMH is equal to ACB; and the angles at K and M being thus equal to BAC and BCA, the remaining angle at L is (I. 30.) equal to that at B, and the two triangles are therefore equiangular.
PROP. XIII. THEOR.
A straight line drawn from the vertex of an equilateral triangle inscribed in a circle to any point in the opposite circumference, is equal to the two chords inflected from the same point to the extremities of the base.
Let ABC be an equilateral triangle inscribed in a circle, and BD, AD, and CD chords drawn from it to a point D in the circumference; BD is equal to AD and CD taken together,
For, make DE equal to DA, and join AE. The angle ADB is (III. 16.) equal to ACB in the same segment, which, being the angle of an equilateral triangle, is equal (I. 30. cor. 1.) to the third part of two right angles. But the triangle ADE being isosceles by construction, the angles DAE, DEA at its base are equal (I. 10.), and each of them is, therefore, equal to half of the remaining
two-thirds of two right-angles, or to one-third part. Con
sequently ADE is likewise an equilateral triangle (I. 11. cor.), and the angle DAE equal to CAB; take CAE from both, and there remains the angle DAC equal to EAB; but the angle ABD is equal to ACD in the same segment. And thus the triangles ADC and AEB have the angles DAC, DCA equal to EAB, EBA, and the interjacent side AC equal to AB; they are consequently equal (I. 20.), and the side DC is equal to EB. But DE was made equal to DA; wherefore DA and DC are together equal to DE and EB, or to DB. **
PROP. XIV. THEOR.
About and in a given square, to circumscribe
and inscribe a circle.
Let ABCD be a square, about which it is required to
circumscribe a circle.
Draw the diagonals AC, DB intersecting each other in O, and, from that point with the distance AO, describe the circle ABCD : This circle will circumscribe the Because the diagonals of the square ABCD are equal and bisect each other, the straight lines OA, OB, OC, and OD are all equal, and consequently the circle described through A passes through the other points B, C, and D. Again, let it be required to inscribe a circle in the square ABCD. From O the intersection of the diagonals and with its distance from the side AD, describe the circle EGHF This circle will touch the square internally. - * * For let fall the perpendiculars OG, OH, and OF (I.6.). And because the straight lines AB, BC, CD, and DA are equal, they are equally distant from the centre O of the exterior circle (III. 10.); wherefore the perpendiculars OE, OG, OH, and OF are all equal, and the interior
circle passes through the points G, H, and F; but (III. 20.)
it likewise touches the sides of the square, since they are perpendicular to the radii drawn from O. Cor. Hence an octagon may be inscribed within a given
square. For let tangents be applied at the points I, K,
L, and M, where the diagonals cut the interior circle. It is evident, that the triangle AOE is equal to DOE, IOP to EOP, and EOZ to MOZ; whence the angles POE and ZOE are equal, being the halves of EOA and EOD, and consequently the triangles PEO and ZEO are equal.
Wherefore PZ, the double of PE, is equal to PQ, the
double of PI; and the angle EZM is, for a like reason, equal to EPI. And, in this manner, all the sides and all the angles about the eight-sided figure PQRSTWYZ are proved to be equal.
PROP. xv. PROB.
In and about a given circle, to inscribe and circumscribe a square.
Let EADB be a circle in which it is required to inscribe a square. Draw the diameter AB, the perpendicular ED through the centre, and join AD, DB, BE, and EA : The inscribed figure ADBE is a square. The angles about the centre C, being right angles, are equal to each other, and are, therefore, subtended by equal chords AD, DB, BE, and AE, but one of the angles ADB, being in a se- G 2 I-I micircle, is (III. 19.) a right angle, so N and consequently ADBE is a square. A aft—B Next, let it be required to circum- N 2% scribe a square about the circle. E J. I Apply tangents FG, GH, HI, and FI at the extremities of the perpendicular diameters: These will form a square. For all the angles of the quadrilateral figure CG, being together equal to four right angles, and those at C, A, and D being each a right angle, the remaining angle at G is also a right angle, CG is a rectangle; and AC being equal to CD, it is likewise a square. In the same manner, CH, CI, and CF are proved to be squares; the sides FG, GH, HI, and IF of the exterior figure, being therefore the doubles of equal lines, are mutually equal, and the angle at G being a right angle, FH is consequently a square. Cor. Hence the circumscribing square is double of the inscribed square, and this again is double of the square described on the radius of the circle,