perpendicular to IG; .. FG = GI. Also HK = HF, .. IK=GF; whence FGGI=IK, and FK is trisected. (96.) If a straight line be divided into any two parts, and upon the whole line and one of the parts, as diameters, semicircles be described; to determine a point in the less diameter, from which if a perpendicular be drawn cutting the circumferences, and the points of intersection and the extremities of the respective diameters be joined, and these lines produced to meet; the parts of them without the semicircles may have a given ratio. Let AB be divided into any two parts in the point C, and on AB, AC let semicircles be described. Take AG: AC the duplicate of the given ratio, and make CD: CB :: AG: GB; D will be the point required. A E F K GL D C From D draw the perpendicular DFE; join BE, CF, and produce them to H; join AE, CI, and from K draw KL parallel to EA. Since CD CB :: AG: GB, comp. and inv. DB: CD:: AB: AG, and since CI is parallel to BE, and KL to EA, whence (Eucl. v. 15.) AB: AG :: AB : CL, consequently AG: AC :: CL: CA, But AG AC is the duplicate of the given ratio, : (97.) If a straight line be divided into any two parts, and from the point of section a perpendicular be erected, which is a mean proportional between one of the parts and the whole line, and a circle described through the extremities of the line and the perpendicular; the whole line, the perpendicular, the aforesaid part, and a perpendicular drawn from its extremity to the circumference will be in continued proportion. per Let AB be divided into any two parts in C, and from C draw the pendicular CD equal to a mean proportional between AB and AC; and through A, B, D let a circle be de B scribed, and draw AE perpendicular to AB; AB, ČD, AC, AE are in continued proportion. = In AB produced take BF-AC. Join FD meeting the circumference in G; join AG, AD, GE. Then because BF AC, .. CF = AB, and CD is a mean proportional between AC and CF, .. ADG is a right angle, whence (Eucl. iii. 21.) AEG is also a right angle, and equal to EAC; . EG is parallel and equal to AB, i. e. to CF; whence (Eucl. i. 33.) EC and GF are equal and parallel, and the angle ACE CFD= ADC, and the triangles AEC, ADC, CDF are similar, = .. (CF=) AB: CD :: CD: CA :: CA: AE. (98.) If the tangents drawn to every two of three unequal circles be produced till they meet; the points of intersection will be in a straight line. Let A, B, C be the centres of the three circles; and let DE, FG, HI be respectively tangents to each of two E R circles, meeting the lines joining the centres in the points P, Q, R; P, Q, R are the points in which two tangents to the circles would intersect. Join PQ, QR; they are in the same straight line. Join AD, AF, BE, BH, CG, CI, and draw BK parallel to PQ. Then BE and AD being perpendicular to PD are parallel, .. AD BE :: AP BP, or AF BH :: AP : BP :: AQ : QK. But (CG) CI: AF :: ex æquo CI: BH: CQ: QK. But CI: BH :: CR: BR, CQ: AQ, ... (Eucl. v. 15.) CR: BR: CQ: QK, also the vertically opposite angles at C are equal, .. the triangles CBK, CQR are similar, and the angle CQR (Eucl. vi. 6.) is equal to BKC, .. CQR and CQP are M together equal to BKC, CQP, i. e. (Eucl. i. 29.) to two right angles, whence (Eucl. i. 14.) PQ and QR are in the same straight line. (99.) If from the extremities of the diameter of a circle any number of chords be drawn, two and two intersecting each other in a perpendicular to that diameter ; the lines joining the extremities of every corresponding two will meet the diameter produced in the same point. From A and B, the extremities of the diameter AB of a semicircle, let AC, BD be drawn intersecting each other in FH, which is perpendicular to AB. Join CD, and produce it to meet BA in P; P is a fixed point, or the line joining the extremities of every other two chords intersecting each other in FH will pass through P. Join BC; and bisect BG in O; and with the centre O, and radius OB, describe a circle HGB, which will circumscribe the quadrilateral figure HGCB. Take E the centre of the semicircle, and join HC, EC. The angle PCE is equal to PCA, ACE together, i. e. to DBA, CAE together; and the angle CHE is equal to ACH, CAH together, i.e. to DBA, CAH together,. PCE CHE, and the angle at E being common, the triangles CEH, CPE are equiangular; = whence EH EC: EC: EP, in which proportion the three first terms being invariable, EP is also, and the point E being fixed, P is also. (100.) If from a given point in the diameter of a semicircle produced, three straight lines be drawn, one of which is inclined at a given angle to the diameter, another touches the semicircle, and the third cuts it, in such a manner, that the distance of the given point from the nearer extremity of the diameter, and the perpendiculars drawn from that extremity on the three aforesaid lines may be proportional; then will the lines, which join the extremities of the diameter and of that part of the cutting line which is within the circle, intersect each other in an angle equal to the given angle. From a given point C, in the diameter AB produced of the semicircle AGB, draw CD inclined at a given H K D F angle to AC, CG touching, and CIH cutting the circle in such a manner that BD, BE, BF being drawn respectively perpendicular to them, CB may be to BD as BE to BF; then if AI, BH be joined, the angle ALH or BLI will be equal to BCD. Join OH, OG; and draw OK perpendicular to HI. Now the angles at E and F being right angles, as also those at G and K, BE is parallel to OG, and BF to OK; ·. (OG=) OH: BE: CO: CB :: OK: BF, .. OH OK :: BE : BF :: BC: BD; also the angle at D is equal to OKH, .. (Eucl. vi. 7.) the triangles OHK, BCD are equiangular, and the angle OHK is equal to BCD. But OHB is equal to OBH, |