ference ADC; and through the point B draw BD perpendicular to AB, meeting the semicircumference at the point D; then will BD be A B a mean proportional to AB, BC. This follows immediately from B. IV, Prop. xxII, Cor. 1. PROPOSITION XXIV. PROBLEM. To construct a rectangle equivalent to a given square, and having the sum of its adjacent sides equal to a given line. D C A Let C be the given square, and AB equal to the sum of the sides of the required rectangle. Upon AB as a diameter, describe the semicircumference ADFB; and draw DF parallel to AB, at a distance equal to a side of the given square, cutting the semicircumference at the points D and F; then through F draw FG perpendicular to AB, and AG and GB will be the sides of the required rectangle. For, FG is a mean proportional between AG and GB (B. IV, Prop. XXII, Cor. 1;) therefore AGXGB= FG". But FG is equal to a side of the given square C; consequently AGX GB the square C. Therefore the given line AB has been divided at the point G, so that the rectangle of the two parts is equal to the square C. Schol. When the side of the given square is greater than half the given line AB, the line DF, drawn as above directed, will not cut the semicircumference; so that in this case the problem would be impossible. (86.) PROBLEM. To find the square and higher powers of a given quantity or magnitude. Draw the straight line AB of indefinite length, on which take the distance AC equal to a unit; through the point C, draw CD perpendicular to AB, and of indefinite length. With A as a centre, and with a radius equal to the given line, describe an arc cutting CD at the point F; through A and F, draw the line AG, of indefinite length; also draw FH perpendicular to AG, HK per H M pendicular to AB, KL perpendicular to AG, LM perpendicular to AB, etc.; then will AH equal the square of AF, AK will equal the cube of AF, AL will equal the fourth power of AF, etc. For, from the manner of construction, it is obvious that the triangles ACF, AFH are right-angled and similar; hence AC=1 AF: AF: AH; therefore AH=AF2. Again, the triangles ACF and AHK are similar, and give 1 AF: AH: AK. Substituting AF2 for AH, we find 1 AF: AF2: AK; therefore AK=AF3. : Comparing the similar triangles AFC and AKL, we have 1 AF: AK: AL. For AK substitute AF3, and we have 1 : AF :: AF AL; therefore AL=AF1. By continuing this operation, we find in succession all the different powers of AF; that is, we find all the terms of the geometrical progression 1, AF, AF2, AF3, AF*, AF", &c. (87.) When the given magnitude is less than a unit, we may use the following construction: Draw the straight line AB of indefinite length; take AC equal to the given magnitude, and through C draw CD at right-angles to AB, of indefinite length. Then with A as a centre, and with a radius equal to a unit, describe an arc cutting CD at the point F: through F draw the line AG, of indefinite length; also draw CH perpendicular to AG, HK perpendicular to AB, KL perpendicular to AG, etc.; then will AH equal the square of AC, AK will equal the cube of AC, AL will equal the fourth power of AC, etc. For, comparing the triangles ACF, ACH, which are similar, we have AF 1 AC :: AC: AH; therefore AH=AC2. Again, comparing the triangles ACF, AHK, we have 1 AC AH : AK. Substituting AC2 for AH, we have 1: AC:: AC2: AK; consequently AK=AC'. And by a similar process we can find the higher powers of AC; that is, we can find all the terms of the geometrical progression. 1, AC, AC2, AC3, AC1, &c. In this case, these terms form a decreasing geometrical progression; while, in the first case, the terms form an increasing geometrical progression. PROPOSITION XXV. THEOREM. If, from the same point without a circle, a tangent and secant line be drawn, the tangent will be a mean proportional between the secant and its external segment. From the point B, let the tangent BA, and the secant BC, be drawn; then will BC: BA :: BA: BD, or BA' BC.BD. For, drawing AD and AC, the triangles ABD and ABC will have the angle at B common; the angle BAD is equal to BCA, (B. III, Prop. VIII;) therefore the two triangles are similar, and we have BC: BA :: BA : BD. which immediately gives BA2=BC.BD. (88.) PROBLEM. In a given straight line AB, required to find a point from which the angle subtended by another given line CD may be the greatest possible. Describe the circle CDF so as to pass through the points C and D, and touch the line AB at F: then F will be the point sought. For, drawing CF and DF, the angle CFD will be measured by half the arc CD. But if we take the point G at the right of F, the angle CGD will be measured by half the difference of the arcs gle CFD is the great est angle which the line CD can subtend from any point in the straight line AB. We will now show how the point F can be found without the aid of the circle. If CD be produced to meet AB at A, we shall have AC a secant line, and AF a tangent; consequently AF is a mean proportional between AC and AD. (B. IV, Prop. xxI.) Hence the distance AF may be found, when the lines AC, AD are given, by the aid of B. IV, Prop. xxi. Remark. So long as the points A, D, and C remain the same, the distance AF will be constant, whatever angle the line BA may make with the line AC. The above problem is equivalent to the following: Through any two points, as C and D, to describe the circumference of a circle which shall touch a given line, as the line AB. From what has already been done, it is obvious we may find the centre of the required circle by the following simple method: Join C and D, and produce CD beyond the line AB until the part AF is equal to AD. On CF describe the semicircumference CGF, and from A draw AG perpendicular to CF, meeting the circumference at G; then with A as a centre, and with AG as a radius, describe an arc cutting the given line AB at H. Through H, perpendicular to AB, draw HL; also through K the mid A D C K H G B |