Since GH is parallel to AD, and GI to DC, the triangles GHI, ADC are similar; but ADC is equilateral, and.. also GHI is equilateral. Also (Eucl. vi. 8. Cor.) ED: EG :: EG: EB, and (Eucl. vi. 2.) ED: EG EA: EH, and. (Eucl. vi. 15.) the triangles EGH, EBA are equal. But GHE=GIE, and BAE=BCE, .. GHI=BAC. ... also (7.) To describe a parallelogram, the area and perimeter of which shall be respectively equal to the area. and perimeter of a given triangle. B E 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. A H For HF-AG=AE, .. HF and AG together are equal to AD, i. e. to AB and BC together; and GF= AH-HC,.. the perimeter of AGFH is equal to the perimeter of ABC; and AGFH is double of a triangle on the base AH and between the same parallels, and .'. is equal to the triangle ABC. (8.) To describe a parallelogram which shall be of given altitude, and equiangular and equal to a given parallelogram. Let ABCD be the given parallelogram, and EF the given altitude. Draw EH and FG at right D angles to FE; and at the H ゴロ point F, in the line GF, make the angle GFI equal to CDA; take FG a fourth proportional to FI, AD and DC; and from G draw GH parallel to FI, meeting EH produced in H; IFGH is the parallelogram required. For its altitude is EF; and the angle GFI=CDA, .. FIH=DAB; whence the parallelograms are equiangular; and they are equal; since the sides about the equal angles are reciprocally proportional (Eucl. vi. 14.). (9.) To describe a square which shall be equal to the sum of any number of given squares. Α Let AB be a side of one of the given squares. From B draw BC perpendicular to AB, and equal to a side of the second square. Join AC; and from C draw CD perpendicular to it, and equal to a side of A the third square. Join AD; and from D draw DE perpendicular to AD, and equal to a side of the fourth. Join AE. The square of AE is equal to the squares of AB, BC, CD, DE. Since the angles ADE, ACD, ABC are right angles, the square of AE is equal to the squares of AD, DE, i. e. to the squares of AC, CD, DE; and .. to the squares of AB, BC, CD, DE. And by proceeding in the same manner whatever b the number of given squares, one equal to their sum may be found. COR. Hence lines may be found, which have the same ratio as the square roots of the natural numbers. (10.) Having given the difference between the diameter and side of a square; to describe the square. D H E B A Let AB be the given difference. Draw AC, AD, each making half a right angle with AB; and complete the square EF. Take AG= the difference between BA and BF. Since the ratio between the side of a square and its diameter is given, that of their difference to the diameter is also given. Take .. AH: AB :: AB: AG; and through H draw HC, HD perpendicular to AC, AD; CD is the square required. For DC being a parallelogram is also (Eucl. i. 46. Cor.) rectangular; and the angle DAH being half a right angle, is equal to DHA, .. DA=DH; whence the sides are equal; and the figure is a square. And since BG=BF, HB=HC; and AB is the difference between the diameter and side. (11.) To divide a circle into any number of concentric equal annuli. Let ABC be the given circle, and O its centre. Draw any radius OA, and divide into the given number DEFG B of equal portions in the points D, E, F, G, &c. On ОA describe a semicircle, and draw the perpendiculars DH, EI, FK, GL, &c. Join OH, OI, &c. and with the centre O and radii OH, 01, &c. let circles be described; they will divide the circle ABC as is required. Since the areas of circles are in the duplicate ratio of their radii; the area of the circle whose radius is OA is to that whose radius is OH in the duplicate ratio of OA : OH, i. e. in the ratio of OA: OD; .. the area of the first annulus will be to the area of the circle whose radius is OD: AD: OD. And in the same manner the area of the second annulus, will be to the area of the circle whose radius is OD, as DE: OD; and since AD=DE, the annuli will be equal. The same may be proved of all the rest. COR. The construction will be nearly the same, if it be required to divide the circle into annuli which shall have a given ratio; by dividing the radius 40 in that proportion. B (12.) In any quadrilateral figure circumscribing a circle, the opposite sides are equal to half the perimeter. Let ABCD be a quadrilateral figure circumscribing the circle EFG; its opposite sides are equal to half the perimeter. For (Eucl. iii. 36. Cor.) AE=AH, and DH=DG, .. AD is equal to AE and H DG together. In the same way BC is equal to BE and GC together, .. AD and BC together are equal to AB and DC together. (13.) If the opposite angles of a quadrilateral figure be equal to two right angles, a circle may be described about it. Let ABCD be a quadrilateral figure, whose opposite angles are equal to two right angles. Join BD; then if a circle be described about the triangle BCD it will pass through A E A. For the angle BCD and the angle in the segment BED, are together equal to two right angles, and .. equal to BCD, BAD; whence BAD is equal to the angle in the segment BED; and ... A must be a point in the circumference; or the circle will be described about ABCD. (14.) A quadrilateral figure may have a circle described about it, if the rectangles contained by the segments of the diagonals be equal. Let ABCD be a quadrilateral figure, the rectangles contained by the segments of whose diagonals are equal, viz. the rectangle AE, EC, equal to BE, ED. B D E Describe a circle about the triangle ABC; if it does not pass through D, let it cut BD in F; then (Eucl. iii. 35.) the rectangle BE, EF, is equal to the rectangle AE, EC, i. e. to the rectangle BE, ED, by the supposition; whence EF is equal to ED, the less to the greater, which is impossible; .. the circle must pass through D. |