the point M draw MN parallel to GH, then will LN be the side of the square sought. For, since MN is parallel to GH, LM: LN:: LG: LH; consequently (Prop. IV. Cor. 1. B. VI.) LM2 : LNo : : LGo : LH2; but, since the triangle LGH is right angled, we have (Prop. XVII. Cor. B. VI.) LG2: LH2 :: GK: KH; hence LM: LN2 :: GK: KH, but, by construction, GK-E and KH=F, also LM-AB; therefore the square described on AB is to that described on LN, as the line E is to the line F. Upon a given straight line ab to construct a polygon similar to a given polygon, ABCDEF. F E D ъв a In the given polygon draw the diagonals AC, AD, AE; and apply the given line ab to AB; making Ab' equal to it:-draw successively b'c', c'd', d'e', ef' respectively parallel to BC, CD, DE, EF; then, from the points ab as centres, with radii equal to Ac, b'c', describe arcs intersecting in c; also, from the centres a, c, with radii equal to Ad', c'd', describe arcs intersecting in d; in like manner, from the centres a, d, with radii equal to Ae', d'e', describe arcs intersecting in e; and, lastly, from the centres a, e, with radii equal to Af, ef", describe ares intersecting in f; then, if the lines bc, cd, de, ef, fa be drawn, the polygon which they form will be similar to that proposed. For the polygon Ab'c'def is similar to the polygon ABCDEF, since they are both composed of the same number of similar triangles, and the polygon abcdef has been made equal to a'b'e'd'e'f'; hence the polygon on ab is similar to that on AB. Scholium. If ab is in the same straight line as AB, or if it is parallel to AB, the construction of this problem will be somewhat simplified. After having divided the proposed polygon into triangles as above, draw from the point a parallels to the diagonals AC, AD, AE; then from b draw be parallel to BC, intersecting the first of these parallels in c; from c draw cd parallel to CD, intersecting the second parallel in d, and so on till the polygon on ab be completed. A polygon being given, to construct a similar polygon that shall be to the former as the line E is to the line F. be the side of the required polygon which is homologous to the side AB; for similar polygons being to each other as the squares of their homologous sides, it follows that the polygon on AB is to the similar polygon on LN as E is to F; therefore it only remains to construct on LN, by the preceding proposition, a polygon similar to that on AB. Two similar polygons being given, to construct another similar polygon which shall be equivalent to either the sum or difference of the former. The method of performing this problem immediately suggests itself from proposition XXII. B. VI. If a right angled triangle be constructed, having its perpendicular sides respectively equal to two homologous sides of the given polygon, the hypothenuse will be the homologous side of a third polygon similar to the former, and equal to their sum; and if two homologous sides of the given polygons be taken, the one for the hypothenuse and the other for a side of a right angled triangle, then the other side of this triangle will be the homologous side of a third polygon similar to the former, and equivalent to their differ ence. Hence, having found in this way a side of the required polygon, the construction is reduced to proposition VIII. PROPOSITION XI. PROBLEM. To construct a polygon, which shall be equivalent to a given polygon P, and similar to another polygon Q. Upon AB, a side of the polygon Q, construct a rectangle AC equivalent to it (Prop. XVI. B. IV.), and on BC describe a rectangle BE equivalent to the other polygon P. Let ab be a mean proportional between AB, BD, then ab will be the side of a polygon similar to Q, and equivalent to P. A B D E P For, if a similar polygon q be constructed on ab, we shall have AB2: ab: Q: q, but, by construction, ab2=AB·BD; therefore AB: AB-BD:: Q: q; consequently (Prop. I. Cor. B. VI.) AB: BD :: Q: q, and, therefore, also AC: BE:: Q: q; but AC=Q; therefore (Prop. IX. Cor. 2. B. V.) BE=q, that is P=q. PROPOSITION XII. PROBLEM. In a given circle to inscribe a triangle similar to a given triangle, abc. Draw a tangent DAE to the circle at any point A in the circumference, and make the angle EAC equal to the angle b, and the angle DAB equal to the angle c. Draw BC, and the triangle ABC will be similar to the triangle abc. For (Prop. XV. B. III.) the angles B E B, C are respectively equal to the angles b, c; therefore the two triangles, being equiangular, are similar. PROPOSITION XIII. PROBLEM. About a given circle to circumscribe a triangle similar to a given triangle, abc. Produce a side bc A Ал of the triangle abc, D For, in the quadrilateral ODBE, the angles O, B are together equivalent to two right angles (Prop. XVII. Cor. 3. B. I.): therefore, since the angle O has been made equal to the exterior angle b, it follows that the angle B is equal to the angle abc. In like manner it will appear that the angle C is equal to the angle acb: hence the triangles ABC, abc are similar. PROPOSITION XIV. PROBLEM. Upon a given base AB to construct an isosceles triangle, having each of the angles at the base double the vertical angle. Produce AB till the rectangle AC.BC may be equal to the square of AB (Prop. XXXII. Cor. B. VI.), then, with the base AB and sides each equal to AC, construct the isosceles triangle DAB, and the angle A will be double the angle D. For make DE=AB, or make AE= BC, and join EB. Then, by construction, AD: AB :: AB: AE, for AE=BC; consequently E the triangles DAB, BAE have a common angle A contained by proportional sides: hence (Prop. XI. B. VI.) they are similar, and, therefore, these triangles are both isosceles, for DAB is isosceles by construction, so that AB EB; but AB-DE; consequently DEEB, and, therefore, the angle D is equal to the angle EBD: hence the exterior angle AEB is equal to double the angle D, but the angle A is equal to the angle AEB; therefore the angle A is double the angle D. Scholium. It is obvious that, in a triangle so constructed, the vertical angle is a fifth part of two right angles, and each angle at the base is two fifths of two right angles, or one fifth of four right angles. PROPOSITION XV. PROBLEM. Upon a given straight line to construct a regular pentagon. Construct, first, upon AB an isosceles triangle DAB, having each of the angles at the base double the vertical angle; and about this triangle circumscribe a circle; then the line AB will be the side of the regular inscribed pentagon. D For, if the radii OA, OB be drawn, the angle O, being double the angle D, will be a fifth part of four right angles; consequently (Prop. XXIII. Cor. 1. B. VI.) the arc AB is the fifth part of the whole circumference, and, therefore, the chord AB is the side of a regular inscribed pentagon. PROPOSITION XVI. PROBLEM. Upon a given straight line AB to construct a regular -hexagon. From the points A, B as centres, with radii equal to AB, describe arcs intersecting in O; and from O, with the same radius, describe a circle, then AB will be the side of the inscribed regular hexagon, as is manifest from proposition VI. book VII. |