Fig. 322.

Fig. 323.

Fig. 324.

Fig. 325.

Fig. 326.

Fig. 327.

Fig. 328.

Fig. 329.

Fig. 319.

Fig. 330.

Q. XXXI. If a circle ADB be described upon the radius AB of another circle, any straight line ADC, drawn from the point A, where they meet, to the circumference of the outer circle, is bisected by the circumference of the inner circle. (105.)

Q. XXXII. If two chords of a given circle intersect each other, the angle of their inclination is equal to half the angle at the centre, which stands on an arc equal to the sum or difference of the arcs intercepted between them, according as they meet within or without the circle.

1. Let AB, CD, cut one another in the point E: and first within the circle ABC; the angle CEA of inclination is equal to half the angle COF, at the centre, standing on an arc equal to the sum of CA and DB. (126.)

2. Let AB, CD, intersect in E, without the circle. (126.)

Q. XXXIII. If two circles ADC, BCE, touch each other in the point C, any straight line ACB, drawn through C, the point of contact, will cut off similar segments.

Draw the diameters CD, CE; and join AD, BE. (128.)

Q. XXXIV. If, through O, the centre of the circle ABC, a circle AOB be described, cutting ABC in A and B, and from A, one of the points of intersection, a straight line AED be drawn, and BE be joined, the part DE, intercepted between the two circumferences, will be equal to the chord BE, drawn from the other point of intersection to the point of meeting of the line with the inner circumference.

Draw the diameter AOC; join BC, BD. (127.)

Q. XXXV. If, from any two points in the circumference of a circle, there be drawn two straight lines to a point in a tangent to that circle, they will make the greatest angle when drawn to the point of contact.

Let A and B be the two points, and CD the tangent at C; join AC, CB; the angle ACB is greater than any other angle ADB formed by lines drawn to any other point D.

Q. XXXVI. If, from any point B in the arc ABC, a line BD be drawn perpendicular to the chord AC, and BF be made equal to BC, and DE to DC, and AF be joined, AF will be equal to AE.

Join FE, EB, FB, BC. (36. 130.)

Q XXXVII. To inscribe a square in a given quadrant of a circle. Let AOB be the given quadrant, whose centre is 0: bisect the angle AOB by the line OC. Draw CE, CD, parallel to OA, OB. DE is a square.

Q. XXXVIII. To describe a circle the circumference of which shall pass through a given point, and touch a given straight line in a given point.

Let AB be the given straight line, C the given point in which the circle is to touch it, and D the point through which it must pass.

Q. XXXIX. To describe a circle the centre of which may be in the perpendicular of a given right angled triangle, and the circumference pass through the right angle, and touch the hypothenuse.

Let EAD be the given right angled triangle, having the angle at A a right angle. Make EC=EA. Join CA, and draw CO at right angles to ED. The circle described with O as a centre, and radius OA, will be the circle required. (45. 48.)

Q. XL. To describe three circles of equal diameters which shall touch each other.

Take any straight line AB, bisect it at D, and erect upon it an equilateral triangle

Q. XLI. If the semicircle ADE be inscribed in the right angled triangle Fig. 331 ABC, so as to touch the hypothenuse BC, in D, and the perpendicular AB, in A, and from E, the extremity of the diameter, a line ED be drawn through the point of contact to meet the perpendicular produced in F; BF, the part produced, will be equal to AB. (152.)

Q. XLII. If, from a point D, taken without a circle ACB, two tangents Fig. 332. DA, DB, be drawn, and a tangent ECF be drawn to another point C in the circumference between them, the sum of the sides of the triangle DEF thus formed, is equal to the sum of the two tangents DA and DB. (152.)

Q. XLIII. If an equilateral triangle be inscribed in a circle, and through Fig. 333. the angular points another be circumscribed, to determine the ratio which they bear to each other.

Let ABC be an equilateral triangle inscribed in the circle about which another DEF is circumscribed, touching the circle in the points A, B, C. (131. 76.)

Q. XLIV. A straight line BD drawn from the vertex B of an equilat- Fig. 334. eral triangle ABC inscribed in a circle, to any point D in the opposite circumference, is equal to the two lines AD, DC, together, which are drawn from the extremities of the base AC to the same point D.

Make DE equal to DA, and join AE. (127. 38.)

Q. XLV. To determine a point within a given triangle, from which lines Fig. 335. drawn to the several angles will divide the triangle into three equal parts.

Let ABC be the given triangle: bisect AB, BC, in E and D; join AD, CE, BF; F is the point required. (170.)

Q. XLVI. If two sides of a trapezoid be parallel, the triangle contain- Fig. 336. ed by either of the other sides and the two straight lines drawn from its extremities to the bisection of the opposite side, is half the trapezoid.

Let ABCD be a trapezoid having the side AB parallel to DC. Let AD be bisected in E; join BE, CE; the triangle BEC is half of the trapezoid. (168.) Through E draw FEG parallel to BC, meeting CD in G, and BA produced in F.

Q. XLVII. If, from any point E in the diagonal AC of the parallelogram Fig. 337. ABCD, straight lines EB, ED, be drawn to the opposite angles, they will cut off equal triangles, viz. the triangles ABE—AED, and BEC=CED. (170.)

Q. XLVIII. The two triangles formed by drawing straight lines from any Fig. 338. point within a parallelogram to the extremities of two opposite sides, are together half of the parallelogram.

Let P be any point within the parallelogram ABCD, from which let lines PA, PB, PC, PD, be drawn to the extremities of the opposite sides; the triangles PAD, PBC, are equal to half the parallelogram, as also the triangles APB, DPC.———Through E draw EPF parallel to AD. (170.)

Q. XLIX. To describe a parallelogram the area and perimeter of which Fig. 339. shall be respectively equal to the area and perimeter of a given triangle.

Let ABC be the given triangle. Produce AB to D, making BD=BC; bisect AD in E; draw BF parallel to AC, and, with the centre A and radius AE, describe a circle cutting BF in G. Join AG, and bisect AC in H. Draw HF parallel to AG. AGFH is the parallelogram required. (168.)

Q. L. If, from A one of the acute angles of the right angled triangle ACD, Fig. 110. a line AB be drawn to the opposite side, the squares of AB and DC are together equal to the squares of AC and BD. (186.)

Fig. 94.

Fig. 340.

Fig. 341.

Fig. 342.

Fig. 343.

Fig. 344.

Fig. 344.

Fig. 345.

Fig. 346.

Fig 347.

Q. LI. In any triangle ABC if a line AD be drawn from the vertex A perpendicular to the base, the difference of the squares of the sides AB and AC is equal to the difference of the squares of the segments of the base, BD and DC. (186.)

Q. LII. If an equilateral triangle ABC, be inscribed in a circle, the square described upon a side thereof is equal to three times the square described upon the radius.

From A draw the diameter AD, and take O the centre, join BD, BO. (126. 186.)

Q. LIII. If two straight lines AC, BD, in a circle, cut each other at right angles, the sums of the squares of the two lines which join their extremities will be equal, viz. the squares of AB and CD will together equal the squares of AD and BC. (186.)

Q. LIV. If, from any point C in the diameter of a semicircle AEB, there be drawn two straight lines CD, CE, to the circumference, one to its point of bisection E, the other perpendicular to the diameter, the squares of these two lines are together double the square of the semi-diameter.

Q. LV. If, from the vertex of an isosceles triangle AOB, a circle be described with a radius less than one of the equal sides, but greater than the perpendicular from O on AB; the parts of the base cut off by it will be equal, viz. AC-BD. Join EF, OC, OD. (199.)

Q. LVI. If three circles touch each other, two of which are equal, the vertical angle of the triangle, formed by joining the points of contact, is equal to either of the angles at the base of the triangle, which is formed by joining their centres.

Let the three circles, whose centres are A, B, C, touch each other in the points D, E, F; and let the two circles whose centres are A and B be equal. Join AB, BC, CA, ED, DF, FE; the angle EDF is equal to either of the angles at A or B. (199.)

Q. LVII. If three equal circles touch each other, to compare the area of the triangle formed by joining their centres with the area of the triangle formed by joining the points of contact.

Let three equal circles, whose centres are A, B, C, touch each other in D, E, F. Join AB, BC, CA, ED, DF, FE. (199.)

Q. LVIII. If a line AB, touching two circles, cut another line CD joining their centres, the segments of the latter will be to each other as the diameters of the circles. (110. 202.)

Q. LIX. If, from a point A without two circles which do not meet each other, two lines AB, AE, be drawn to their centres, which have the same ratio that their radii have, the angle contained by tangents AC, AD, and AF, AG, drawn from the point A towards the same parts of the two circumferences, will be equal to the angle contained by the lines drawn to the centre, viz. CAD and FAG will each be equal to BAE. (208.)

Q LX. If two circles touch each other externally, and parallel diameters be drawn, the straight line joining the extremities of these diameters will pass through the point of contact.

Let ABG, DGE, be two circles touching each other externally in the point G; and let AB, DE, be parallel diameters; join AE; AE will pass through G. Join O, C, the centres of the circles; OC will pass through G (117); let it meet AE in F. (202. IV.)

Q. LXI. If two circles touch each other, and also touch a straight line, Fig. 348. the part of the line between the points of contact is a mean proportional between the diameters of the circles.

Let AEB, CED, be two circles touching each other in E and a straight line AC in A and C; draw the diameters AB, CD; AC is a mean proportional between AB and CD. Join AD, BC; these lines (Q. LX) pass through the point of contact. (202.)

Q. LXII. If three straight lines, drawn from the same point and in the Fig. 349. same direction, be in continued proportion, and from that point also a line equal to the mean proportional be inclined at any angle, the lines joining the extremity of this line and the lines in proportion, will contain equal angles.

Let AB: AC:: AC: AD, and from A let AE be drawn equal to AC inclined at any angle to AB; join EB, EC, ED: the angle BEC is equal to the angle CED. (208.)

Q. LXIII. If, from the extremities and the point of bisection C of Fig. 350. the arc AB, lines AD, BD, CD, be drawn to any point D of the opposite circumference, and AB, AC, be joined, these lines will give the proportion AD+DB: DC::AB: AC. Draw AE parallel to CD, and let it meet BD produced in E. (76. 202. 48.)

Q. LXIV. In any right angled triangle ABC, the straight line CD, joining Fig. 351 the right angle and the point of bisection of the hypothenuse AB, is equal to half the hypothenuse. (196.)

Q. LXV. If the points of bisection D, E, F, of the sides of the triangle Fig. 352. ABC be joined, the triangle DEF so formed is one fourth of the triangle ABC.

Q. LXVI. If, from B one of the equal angles of the isosceles triangle, a Fig. 353. perpendicular BD be drawn to the opposite side, the part BF intercepted by a perpendicular AE drawn from the vertex to the base will have to one of the equal sides the same ratio that the segment BE of the base has to the perpendicular AE; that is, BF: AC:: BE:AE.

Q._LXVII. If, from any point D in the base of an isosceles triangle ABC, Fig. 354. lines DE, DF, be drawn to the opposite sides, making equal angles with the base, the triangles AED, CDF, formed by these lines, the segments of the base and the lines AE, CF, joining the intersection of the sides and the angles opposite, will be equal. (202. 217.)

Q. LXVIII. If, in two triangles, the vertical angle of the one be equal to Fig. 355. that of the other, and one other angle of the former be equal to the exterior angle at the base of the latter, the sides about the third angle of the former shall be proportional to those about the interior angle at the base of the latter. Let ABC, DEF, be two triangles having the angle BAC equal to EDF, and ABC equal to the exterior angle DFG, made by producing the side EF; then AC: CB:: DE: FE. At the point D in the line FD, make the angle FDG equal to the angle EDF or BAC, and meeting EF produced in G. (201.)

Q. LXIX. If, from the three angles of a triangle, lincs be drawn to the Fig. 356. points of bisection of the opposite sides, these lines intersect each other in the same point.

Let the sides o ne triangle ABC be bisected in D, E, F. Join AE, CD, meeting each otner in G. Join BG, GF; BGF is a straight line. Join EF meeting CD in H. (202. 208.)

Q. LXX. The three straight lines which bisect the three angles of a Fig. 356. triangle meet in the same point.

Fig. 357.

Fig. 358.

Fig. 359.

Fig. 360.

In the triangle ABC let the angles at A and C be bisected by the lines AE, CD, and through G their point of intersection draw BGF; it bisects the angle at B. (201.)

Q. LXXI. If, from any point D in the side AB of the triangle ABC, two lines DC, DE, be drawn, the one to the opposite angle and the other parallel to the base, and if DC intersect in G, a line BF, drawn from the vertex to the middle of the base, A, G, and E, are in the same straight line.

Q. LXXII. If, in a parallelogram ABCD, a line AF be drawn from the angle A to the middle of the opposite side DC, the segment DH of the diagonal made by this line will be one third of the whole diagonal.

Q. LXXIII. If, from any angle A of a rectangular parallelogram ABCD, a line AE be drawn to the opposite side, and from the adjacent angle B of the trapezoid thus formed, a line BF be drawn perpendicular to the former, the rectangle contained by the two lines AE, BF, is equal to the given parallel

ogram.

QLXXIV. If, through any point D within the triangle ABC, HG, EF, IK, be drawn parallel to the sides, then IDXDG×DF=ĚD×DKXDÍ. Fig. 361.

Fig. 362.

Fig. 363.

Fig. 364.

Fig. 365.

Fig. 366.

Q. LXXV. If, from any point C in the diameter BA produced, a tangent CD be drawn to the circle, a perpendicular DE from the point of contact to the diameter will divide it into segments which give the proportion AE: EB:: AC: CB.

Take O the centre of the circle, and join DO. (215.)

Q. LXXVI. If, from the extremity B of the diameter AB, a line BC be drawn in the semicircle equal to the radius OB, and from the centre a perpendicular OD be let fall upon it and produced to the circumference, it will be a mean proportional between DB and DA, lines drawn from the intersection D to the extremities of the diameter.

Join DC. (126.)

Q. LXXVII. A straight line AB being divided in two given points C and D, to determine a third point F, such that its distances from the extremities A and B may be proportional to its distances from the given points. (215.)

Q. LXXVIII. To divide a straight line AB into two parts such that the rectangle contained by them may be equal to the square of their difference. Upon AB describe a semicircle ADB. From B draw BC at right angles, and equal to AB. Take O, the centre, and join OC, and from D draw DE perpendicular to AB: AB is divided in the point E, as was required. (215.)

Q. LXXIX. To determine two lines such that the sum of their squares may be equal to a given square, and their rectangle equal to a given rectangle. Let AB be equal to a side of the given square. Upon it describe a semicircle ADB; and from B draw BC perpendicular to AB and equal to a fourth proportional to AB, and the sides of the given rectangle. From C draw CD parallel to BA. Join AD, DB; they are the lines required.

Q. LXXX. Through a given point to draw a line terminating in two lines. given in position, so that the rectangle contained by the two parts may be equal to a given rectangle.

Let AB, CD, be the lines given in position, E the given point; from E draw EF perpendicular to AB, and produce FE to G, so that the rectangle FE, EG, may be equal to the given rectangle. On EG describe a circle cutting CD in H. Join HE and produce it to A; AH is the line required. Join GĚ.