Then by the conditions a x :: x: α-x, But this is a quadratic equation in x, solving which we obtain I operations here represented. For if AB = a, HB = a, and 2 The negative sign before the radical corresponds to the solution of a problem more general than the geometrical problem as stated above, and the consideration of it must be deferred. PROBLEM 5. To construct a triangle similar to a given triangle, on a straight line which is to be homologous to a given side of the triangle. Let ABC be the given triangle, DE the given straight line which is to be homologous to AB. Construction. At D and E draw lines making with DE angles equal to the angles A and B, and let these lines meet in F. Then DEF is the triangle required. Α B D E - Proof. For the triangle DEF is by construction equiangular to the triangle ABC, and therefore it is similar to it. COR. Hence a polygon can be constructed similar to any given polygon, on a straight line which is to be homologous to a given side of the polygon. For the given polygon can be divided into triangles. PROBLEM 6. To make a square which shall be to a given square in a given ratio. Let AB be the side of the given square, AB : BC in the given ratio. Construction. On AC describe a semicircle, and draw BD perpendicular to AC, to meet the semicircle in D. BD is the side of the square required. Proof. Since AB : BD :: BD : BC, therefore that is AB2: BD2 :: AB : BC, AB: BD in the given ratio. COR. The same construction and proof are applicable to any polygon. For similar polygons are to one another as the squares on their homologous sides. PROBLEM 7. To inscribe a regular decagon in a given circle. We shall solve this by the method of analysis and synthesis. Analysis. The radii to two consecutive angular points must include an angle equal toth of 4 right angles, or I to 5 ΙΟ th of 2 right angles. That is, if OA, OB are such radii, the angle at O must be I th of two right angles, and there 5 fore the angles at A and B must each be angles, since the three angles are together equal to 2 right angles. Bisect the angle OBA by the line BC. Then OBC and CBA each are equal to AOB; and therefore CB-CO, and BCA is double of COB, and therefore = BAC; therefore AB = BC = CO. B But since OBA is bisected by BC therefore OB: AB :: OC : CA; OA OC: OC: CA, or OA is divided in extreme and mean ratio in C. Hence the construction follows. Synthesis. Take OA any radius of the circle; divide it in extreme and mean ratio in C so that OA OC: OC: CA. Place AB as a chord of the circle equal to OC. Join BO, CA. and Then AB is a side of a decagon inscribed in the circle. therefore and therefore CB bisects the angle OBA; and the angle OBA is double of ABC. And again since the triangles OAB, BAC have the angle at A common, and have the sides about the common angle proportionals, viz. OA : AB :: AB : AC; Therefore these triangles are similar and equiangular; therefore the angle ABC is equal to the angle AOB. But OBA is double of ABC, and therefore each of the angles OBA, OAB is double of AOB. But the three angles OBA, OAB, AOB together equal 2 right angles; therefore OAB =th of 2 right angles, or 5 Hence ten chords equal to AB could be placed round the circumference of the circle, and thus a regular decagon would be inscribed. COR. I. Hence a regular pentagon may be inscribed in a circle by joining the alternate angular points of an inscribed decagon. COR. 2. Hence regular polygons of 20, 40, 80 sides can be constructed. PROBLEM 8. To inscribe a regular quindecagon in a given circle. Let ABC be the given circle, in which it is required to inscribe a regular quindecagon. |