and PQ a chord to the greater, touching the less in T, then PT = TQ. For if OT be drawn, it will be perpendicular to PQ (III. 18); and since O is the centre of the circle MPN, and OT perpendicular to the chord PQ, it bisects the chord (III. 3); hence PT TQ. EXERCISE XIII.-THEOREM. If any number of circles intersect a given circle, and pass through two given points, the lines joining the intersections of each circle will all meet in the same point. Let A, B be two given points, and CDFE a given circle; then if any number of circles ABC, ABE be described through A, B, cutting the circle CDFE, then the lines CD, EF, produced, will meet AB in the same point P. For let CD meet AB, when both are produced, in the point P, and from P draw a line PFE, cutting the circle CDFE in E and F; then the circle described = E through A, B, and E, will cut PE in F, the point in which it cuts CDFE. For if not, let it cut PE in another point F' (not shewn in the figure); then as PC, PE are secants of the circle CDFE, therefore (III. 37) PE PF PC PD PA PB; but PA, PF'E, being secants of the circle ABF'E, therefore PA PB = PE PF'; hence PE PF = PE PF'; and consequently PF = PF', or F' coincides with F. Wherefore the line joining the points of section E, F, meets AB in the same point P in which CD meets it; and the same can be proved of any other circle passing through A and B, and cutting the circle CDFE. = A D B P COR.-Hence if the rectangles PA PB and PE PF are equal, a circle can be described through A, B, E, and F. EXERCISE XIV.-PROBLEM. Through two given points, to describe a circle touching a given circle. Let A, B be the given points, and CDT the given circle, to describe a circle through A and B, to touch CDT. Let any circle passing through A and B cut the given circle in C and D; draw CD and AB, and produce them to meet in P; from P draw the tangents PT, PT' to the given circle CDT; then if a circle be described passing through A, B, and T, or through A, B, and T', it will touch the given circle either externally or internally. = For since PT is a tangent to the given circle CTD, therefore PT2 PC PD (III. 36); but A, B, C, D being, by hypothesis, points in the circumference of a circle, hence PC PD = PA PB; wherefore PA PB = PT2, and consequently (III. 37) PT is also a tangent to the circle ABT. Since PT is a common tangent, the circles must touch in T; for the tangent lies between them, so that they can meet only in the point T (III. 11 and 12). Α. T B P If a circle be described through A, B, and T', it could be similarly proved that it would touch the given circle in T'. EXERCISE XV.-THEOREM. If two chords in a circle intersect each other perpendicularly, the sum of the squares of their four segments is equal to the square of the diameter. Let ABDC be a circle, and AB, CE two perpendicular chords in it, then the sum of the squares of the segments AF, FC, EF, FB is equal to the square of the diameter. D For draw the diameter ED, and join CD, AC, DB, and EB. Then angle ECD is a right angle (III. 31), and therefore = EFB; consequently (I. 29) CD is parallel to AB; and hence (9th Ex.) the arc AC = DB; wherefore (III. 26) the chord AC = DB. Also the angle EBD in a semicircle is a right A angle (III. 31). Now (I. 47) AF2 + FC2= AC2 DB2; and EF2 + FB2 EB2; also EB2 + DB2 = DE2; consequently, adding equals to equals, AF2 + FC2 + EF2 + FB2 = DE2. = EXERCISE XVI.-THEOREM. E F Perpendiculars from the extremities of a diameter of a circle, upon any chord, cut off equal segments. Let ABDC be a circle, AB a diameter, CD a chord, and AE, BF perpendiculars on the chord; then the segment CE = DF. For draw GH through the centre O, parallel to CD, and OI perpendicular to EF. A Then because AE, OI are perpendicular to EF, they are parallel (I. 28); and EL, GO are parallel by construction; consequently E G F B H GI is a parallelogram, and it is rectangular (I. 46, Cor.). It is similarly shewn that IH is a rectangle; wherefore GO EI, and = at O are equal, the OH IF. Now, in the triangles AOG, BOH, the vertical angles alternate angles at A and B are also equal, OB; consequently (I. 26) the triangles are hence GO = OH; wherefore EI = IF, and therefore the remainder CE = DF. and the side AO every way equal; (III. 3) CI = ID; EXERCISE XVII.-PROBLEM. Given the vertical angle, the base and altitude of a triangle, to construct it. Let AB be the base, V the vertical angle, and H the altitude of the triangle, to construct it. On the base AB describe a segment of a circle ABDE, containing an angle equal to the given angle V; and draw AC perpendicular to AB, and make AC = H; through C draw CD parallel to the base, and CE let it cut the arc in D and E; join AD and DB, and ADB is the required triangle. H For the vertex of the triangle must lie in the line CD (I. 34, Cor.); it must also evidently lie A in the arc of the segment (III. 21); hence it may be either of the points D or E. Join AD and DB, and ADB is one of the required triangles; for it has the given base, the given height, and vertical angle. V ہے B By joining AE and BE, it can be similarly shewn that the triangle ABE fulfils the three required conditions of having its base, altitude, and vertical angle of the given magnitudes. The two triangles ADB, ABE are evidently equal in every respect, though they are in reverse positions. They could not be made to coincide, without first conceiving one of them to be turned round a line perpendicular to the extremity of the base, as an axis, till it would describe half a revolution, and thus again come into the same plane, though in a reverse position to its present, and then be drawn along the length of the base. EXERCISE XVIII.-THEOREM. In a circle, the sum of the squares of two lines drawn from the extremities of a chord, to any point in a diameter parallel to it, is equal to the sum of the squares of the segments of the diameter. Let ABDC be a circle, of which AB, CD A are a diameter and a chord parallel to it; then if E be any point in the diameter, CE2+ED2 = AE2 + EB2. B O E For take O the centre, and draw OF perpendicular to CD, and join EF and OC. Then (III. 3) CD is bisected in F; OF is perpendicular to AB (I. 29); and hence (II. A) CE2 + ÉD2 = 2CF2 = = +2EF2. Again, 2EF2 = 20E2 + 20F2 (I. 47); consequently 2CF2+2EF2 2CF2+20F2+20E2. But 2CF2 + 20F2 2002 =2AO2 (I. 47); therefore 2CF2 + 2EF2 AE2 + EB2 (II. 9). Hence CE2 + ED2 = AE2+ EB2. = EXERCISE XIX.-PROBLEM. 2A02 + 20E2 Given the vertical angle, the base, and the sum of the sides of a triangle, to construct it. Let AB be the base, V the vertical angle, and S the sum of the sides, to construct the triangle. On the base AB describe a segment of a circle ACB, containing an angle equal to V, and another arc, of which BD is a portion, containing an angle = V; then with a radius AE: = S, and centre A cut BD in D; join AD and BF, and ABF is the required triangle. A For join BD; then the angle AFB = twice ADB by construction; but AFB ADB + DBF (I. 32); hence ADB + DBF = 2ADB; and therefore DBF = FDB, and consequently DF = BF. But AF + FD = AD = AE = S; therefore also AF + FB S. The triangle ABF, therefore, has the given base, the given sum of the sides, and the given vertical angle; it is, therefore, the required triangle. S Had the difference of the sides been given instead of the sum, then, instead of describing the segment of which BD is a part, it would have been necessary to describe a segment on AB containing an angle equal to 2V; and then from A as centre, with a radius equal to the difference, to cut this latter arc in a point, say D; then the remaining construction and proof would have been analogous to that above. EXERCISE XX.-THEOREM. M If the points of contact of two tangents to a circle be joined, any secant drawn from their intersection is divided into three such segments, that the rectangle under the secant and its P middle segment is equal to that under its extreme segments. Let PMN be a circle, TM, TN two tangents, MN a chord joining the points of contact, and PT a secant; then = N R For (III. 37, Cor.) MT TN, and hence MNT is an isosceles triangle; consequently (II. D). but hence MT2 = QT2 + MQ • QN = QT2 + PQ QR (III. 35); Or, since therefore QT2 = QT · QR + QT · TR (II. 2); QT QR+QT·TR+PQ • QR = PQ • TR+QT TR; and taking QTTR from both these equals, there remains QT · QR + PQ • QR = PQ • RT. Or (II. 1), PT QR PQ⚫RT. EXERCISE XXI.-THEOREM. If the opposite sides of a quadrilateral inscribed in a circle be produced to meet, the square of the line joining the points of concourse is equal to the sum of the squares of the two tangents from these points. Let CDEF be a quadrilateral inscribed in a circle, and let the opposite sides produced meet in A and B, then AB2T2+T'2, where T, T denote the lengths of the tangents AT, BT', drawn from A and B to the circle. Divide AB in G, so that AB BG A T G D E B T2. This can be done by drawing from B a secant to the circle CDEF equal to BA; then its external segment will be BG. Then since T/2 = BD BE (III. 36), hence AB⚫BG= DB BE, and consequently a circle may be described through the points A, G, E, and D; hence angle AGD = DEF (III. 22); wherefore (I. 13, Cor., and III. 22) since angle AGD + BGD = two right angles, and DEF+ DCF two right angles, therefore angle BGD = BCD, and consequently a circle may be described through the points B, G, C, and. D; and consequently AB AG = AD ACT2; but AB BG = T2, whence AB AG + AB· BG = T2 + T'2. = But AB AG + AB·BG = AB2 (II. 2), and therefore AB2 = T2 + T'2. EXERCISE XXII.-PROBLEM. Two parallel chords in a circle are respectively six and eight inches in length, and are one inch apart; find the diameter of the circle. In the figure to Ex. 6, let DE: 6, AB = 8, and FG 1; join CE, then CE = CB = half the diameter. In the right-angled triangles CGE and CFB, CE2= CG2 + GE (I. 47) and CB2 = CF2 + FB2; but since CE CB their |