THEO. XXI. Equiangular triangles ABC, DEF, are to one another in a duplicate proportion of their homologous or like sides; or as the squares AK and DM of their homologous sides. Let the perpendiculars CG and FH be drawn, as well as the diagonals BI and EL. The perpendiculars make the triangles ACG and DFH equiangular, and therefore similar A B D H K = FDH and the (by Theo. 15.) for because the angle CAG = Therefore GC FH:: (AC: DF) :: AB: DE, or which is the same thing GC: AB:: FH: DE, for FH multiplied by AB AB multiplied by FH. = (By Theo. 18) ABC: ABI::(CG: AI, or AB :: FH: DE, or DL) :: DFE: DLE, therefore ABC: ABI:: DFE: DLE, or ABC: AK::DFE: DM, for AK is double the triangle ABI, and DM double the triangle DEL, (by Cor. 2. Theo. 11.) Q. E. D. THEO. XXII. Like polygons ABCDE, abcde, are in a duplicate proportion to that of the sides AB, ab, which are between the equal angles A and B, and a and b, or as the squares of the sides AB, ab. Draw AD, AC, ad, ac. A B equal angles A, and a, we take the equal ones EAD + BAC ead bac, the remaining angle DAC dac, and if from the equal angles D and d, EDA eda be taken, we have ADC taken BCA = adc: and in like manner if from C and c be bca, we have ACD acd; and so the respective angles in every triangle will be equal to those in the other. (By Theo. 21.) ABC: abc:: the square of AC to the square of ac, and also ADC: adc:: the square of AC to the square of ac; therefore from equality of proportions ABC: abc:: ADC: adc, in like manner we may show that ADC: adc: EAD: ead: therefore it will be as one antecedent, is to one consequent; so are all the antecedents, to all the consequents. That is ABC: abc as the sum of the three triangles in the first polygon, is to the sum of those in the last. Or ABC will be to abc, as polygon to polygon. The proportion of ABC to abc (by the foregoing Theo.) is as the square of AB is to the square of ab, but the proportion of polygon to polygon is as ABC to abc as now shewn: therefore the proportion of polygon to polygon is as the square of AB, to the square of ab. Q. E. D. THEO. XXIII. Let DHB be a quadrant of a circle described by the radius CB: HB an arc of it, and DH its complement; HL or FC the sine; FH or CL its co-sine; BK its tangent; DI its cotangent; and CK its secant; and CI its co-secant. 1. The co-sine of an arc, is to the sine, as radius is to the tangent. 2. Radius is to the tangent of an arc, as the co-sine of it is to the sine. 3. The sine of an arc is to its co-sine, as radius to its cotangent. 4. Or radius is to the co tangent of an arc, as its sine to its co-sine. F K B 5. The co-tangent of an arc is to radius, as radius to the tangent. 6. The co-sine of an arc is to radius, as radius is to the secant. 7. The sine of an arc is to radius, as the tangent is to the secant. The triangles CLH, and CBK being similar, (by Theo. 15.) 1. CL: LH:: CB: BK. 2. Or, CB: BK:: CL: LH. The triangles CFH, and CDI, being similar. 3. CF (or LH): FH :: CD : DI. 4. CD: DI:: CF, (or LH): FH. = The triangles CDI and CKB are similar; for the angle CID KCB, being alternate ones (by part 2. Theo. 3.) the lines CB and DI being parallel: The angle CDI = CBK being both right angles, and consequently the angle DCI CKB, wherefore, 5. DI: CD::CB: BK. = And again, making use of the similar triangles CLH, and CBK. 6. CL: CB: CH: CK. 7. HL: CH: BK: CK. CHAPTER III. PRACTICAL GEOMETRY. GEOMETRICAL PROBLEMS. PROBLEM I. To bisect a given right line AB, that is, to cut it at right angles by another right line CD. 1. From the point A as a centre, and with any radius greater than half the length of the given line, describe the A arcs a and b, above and below the said line. 2. With the same radius and from B as a centre, describe the arcs c and d, intersecting the former. 3. Through the points of intersection, draw the line DE, which will divide the given line into two equal parts in the point C, as required. NOTE.-CD is a perpendicular raised in the middle of the line AB, therefore the four angles about the point C are right angles: also by this problem, a circle can be divided into four equal parts called Quadrants. E PROB. II. To draw a line parallel to a given line AB. Case I. When the parallel line is to be at a given distance. 1. Take two points c and d in the given line; and from the G points c and d as centres, and the given distance for a radius, describe the arcs e and f. 2. Draw the line GH, touching 온 d D B both arcs without cutting them, and it will be parallel to AB as required. Case II. When the parallel line is to pass through a given point C. To divide a given line AB into any number of equal parts. 1. Draw the line AC, and make an angle at B, equal to the given angle at A. 2. With any convenient distance, set off the number of parts required (suppose four) from A towards C, A 4 C 1 and also the same parts from B towards D. 3. Draw the lines (A.4), (1.3), (2.2), (3.1), (4.B), which will divide the line AB, as required. NOTE. This operation not only divides AB, but also AC, or BD, into four equal parts. |