« PreviousContinue »
of any other chord passing through the same point ; the line joining their intersections of the circle will be parallel to the
On opposite sides of any point C of the chord AB of the circle ABG let two lines CD and CE be taken, such that the rectangle DC, CE may be equal to the rectangle AC, CB; and through C let any chord GCF be drawn, and DF, GE joined, meeting the circumference in H and I. Join HI; it will be parallel to AB.
Since the rectangle DC, CE is equal to the rectangle AC, CB, i.e. to the rectangle GC, CF,
.. DC : CF :: GC : CE, .. the triangles DCF, GCE are equiangular, and the angle FDC is equal to CGE or FHI in the same seg. ment; .. HI is parallel to AB.
(26.) If from two points without a circle two tangents be drawn, the sum of the squares of which is equal to the square of the line joining those points ; and
from one of them a line be drawn cutting the circle, and two lines from the other point to the intersections with the circumference; the points in which these two lines cut the circle, are in the same straight line with the former point.
From A and B two points without the circle CDE let tangents AC, BD be drawn, such that the sum of their squares may be equal to the square of AB. If
from A any line AFE be drawn, and BF, BE joined; the points A, H, G will be in a straight line.
In AB take a point I, so that the rectangle AB, BI may be equal to the square of BD. Join IF, IH, AH, HG. Then the rectangles AB, AI and AB, BI are together equal to the square of AB, i. e. to the squares of AC, BD; .. the rectangle AB, AI is equal to the square of AC or to the rectangle AF, AE;
.. AF : AB :: AI : AE,
and .. (Eucl. vi. 6.) the angle AIF is equal to AEB, whence also FIB=FHG. Now the rectangle BI, BA being equal to the square of BD, or to the rectangle BH, BF,
.. BF : BA :: BI : BH, and .. the angle AHB is equal to FIB, or FHG; and BHF is a straight line, .. AHG is a straight line.
(27.) If from the vertex of a triangle there be drawn a line to any point in the base, from which point lines are drawn parallel to the sides; the sum of the rectangles of each side and its segment adjacent to the vertex will be equal to the square of the line drawn from the vertex together with the rectangle contained by the segments of the base.
From the vertex A of the triangle ABC let a line AD be drawn to any point D in the base; from which let DF, DE be drawn parallel respectively to AB, AC; the
rectangles BA, AE, and CA, AF will together be equal to the square of AD, and the rectangle BD, DC together.
About ABC let a circle be described ; and let AD meet the circle in G. Join BG, GC. From E draw EH, making the angle AHE equal to ABG. Produce AB to 1. Because the angles AHE, ABG are equal, the points E, B, G, H are in the circumference of a circle, ... the rectangle BA, AE is equal to the rectangle GA, AH; and the angle EHD will also be equal to GBI, i. e. to ACG. And because AC, DE are parallel, the angle EDH is equal to GAC; hence the triangles EDH, GAC are equiangular,
and .:. AC : AG :: DH : DE, and the rectangle AG, DH is equal to the rectangle AC, DE, i.e. to the rectangle AC, AF. And because the rectangle BA, AE is equal to the rectangle GA, AH and the rectangle CA, AF to the rectangle AG, DH, .. the rectangles BA, AE and CA, AF are together equal to the rectangles GA, AH, and GA, DH, i. e. to GA, AD or to the rectangle AD, DG together with the square of AD; or to the rectangle BD, DC together with the square of AD.
(28.) If on the chord of a quadrantal arc a semicircle be described ; the area of the lune so formed will be equal to the area of the triangle formed by the chord and terminating radii of the quadrant.
Let ABO be a quadrant, on the chord of which let a semicircle ADB be described ; the lune ADBE is equal to the triangle ABO.
Since circles are as the squares of their radii, the quadrant AEBO : ADC :: A02 : AC :: 2 : 1, :.
.: , the quadrant AEBO is equal to the semicircle ADB; and taking away the part AEBC, the lune ADBE is equal to the triangle ABO.
(29.) If from the extremities of the side of a square circles be described with radii equal the former to the side, and the latter to the diagonal of the square ; the area of the lune so formed will be equal to the area of the square.
From D and C the extremities of DC the side of a square, with radii DB and CB, let circles be described, cutting each other again in E; the area of the lune BFE is equal to the square AC.
. Join CE, ED. Since BC is equal to CE, and CD is common, and BD is equal to DE, .. the angles BCE, ECD are equal; whence BE is a straight line; also the angles BDC, EDC are equal; and BDC being half a right angle, BDE is a right angle; :: the arc BE is a quadrant; .. the lune BFE is equal to the triangle BDE (viii. 28.) i. e. to the square AC.
(30.) If on the sides of a triangle inscribed in a semicircle, semicircles be described, the tuo lunes formed thereby will together be equal to the area of the triangle.
Let ABC be a triangle inscribed in a semicircle. On AB, BC let semicircles ADB, BEC be described ;
the lunes ADBF, BGCE are together equal to the triangle ABC.
Since the areas of circles are as the squares of their diameters, the semicircle ABC : ADB :: AC : AB', and ABC : BEC :: AC2 : BC
:. ABC' : ADB+BEC :: AC? : AB'+BC, ine. in a ratio of equality, ... ABC = ADB + from these equals take away the segments AFB, BGC, , and the triangle ABC=AFBD+BGCE.
(31.) If on the two longer sides of a rectangular parallelogram as diameters, two semicircles be described towards the same parts; the figure contained by the two remaining sides of the parallelogram and the two circumferences shall be equal to the parallelogram.
Let ABCD be a rectangular parallelogram, on the sides AB, DC of which let semicircles AEB, DFC be described; the figure DAEBCHFG is equal to ABCD.
Since AB = DC, the semicircles are equal ; from each of which take away FGH, and AGFHBE = DGHC; if to these equals be added ADG and BHC, the whole ADGFHCBE will be equal to the whole ABCD.
(32.) If two points be taken at equal distances from the extremities of a quadrant,
a quadrant, and perpendiculars be