21. The data as above to draw two lines from the two given points, meeting in the given line, and making equal angles with it. theorems to what is given (or may be obtained), and is the natural process of discovery or invention. The required result having been thus obtained by analysis, or resolution, the demonstration of its correctness is made by synthesis, or composition; the order in which is the reverse of the former, and carries us forward from the data, by means of the truths on which the result depends, to the result itself. Analysis is, then, the method of discovery, synthesis of demonstration after the discovery is made. The one has for its object to find unknown truths, the other to prove known ones. Analysis and synthesis are both of them applicable to theorems as well as problems. In submitting a problem to analysis, its solution, in the first instance, is assumed; and from this assumption a series of consequences are drawn, until at length something is found which can be done upon established principles. In the synthesis, or solution, beginning with the construction indicated by the final result of the analysis, the process ends with the performance of what was required by the problem, and is the first step of the analysis. In When a theorem is submitted to analysis, the thing to be determined is whether the statement expressed by it be true or not. the analysis this statement is, in the first instance, assumed to be true, and a series of consequences deduced from it, until some result is obtained, which either is an established or admitted truth, or contradicts an established or admitted truth. If the former, the theorem may be proved by retracing the steps of the investigation, commencing with the final result and concluding with the proposed theorem. But if the final result contradict an established truth, the proposed theorem must be false, since it leads to a false conclusion. We give a specimen of the analytic investigation of a problem be low. Of synthesis the student has already had specimens in all the preceding theorems, and will find others in the problems which follow the remaining theorems of plane geometry. Specimen of the Analysis of a Problem. Given two angles, and the sum of the three sides of a triangle, to construct it. Suppose it done, and that ABC is the triangle sought. Produce BC till CD = CA and BE = BA; E join EA, DA; then the triangles B ACD and ABE being isosceles, the angle ABC = A C twice the angle Hence the following construction: at the extremities of a line ED, equal to the given sum of the three sides, draw lines making, with this, angles each equal to half one of the given angles, and from the point A, where they meet, draw lines making angles with AE and AD, respectively equal to the angles E and D; ABC will be the triangle required. The demonstration synthetically would be as follows: The angle EAB being: the angle E, the triangle is isosceles, and 22. The same when the two given points are on opposite sides of the line. 23. When every side of a polygon is produced out, prove that the sum of the outward angles is equal to four right angles. 24. Show that in an isosceles triangle the square of the line drawn from the vertex to any point of the base, together with the rectangle of the segments of the base, is equal to the square of one of the equal sides of the triangle. 25. Prove that the square of a line is equal to the square of its projection on another line added to the square of the difference of the perpendiculars which determine this projection. 26. Prove that the sum of the squares on two lines, together with twice their rectangle, is equal to the square on their sum. 27. That the square on the difference of two lines is equal to the sum of their squares minus twice their rectangle. 28. To construct a quadrangle when three sides, one angle, and the sum of two other angles are given. 29. When three angles and two opposite sides. 30. Prove that two parallelograms are equal when they have two sides and the included angle equal. 31. Prove that the greater diagonal of a parallelogram is opposite the greater angle. 32. Prove that two rhombi are equal when a side and angle of the one are equal to the same in the other. 33. That if the diagonals of a quadrilateral bisect each other at right angles, the figure will be a rhombus. 34. That the diagonals of a rectangle are equal; and the converse. 35. Prove that the line joining the middle points of the inclined sides of a trapezoid is parallel to the bases, and that it is equal to half the sum of the bases. 36. Prove that two convex polygons are identical: 1°. When they have the same vertices. 20. When one side in each equal, and the distances of the corresponding vertices from its extremities equal. 30. When composed of the same number of equal triangles, similarly placed. 40. When they have all their sides equal and all their augles but two. 50. When all their sides but one and all their angles but one. 37. Prove that there can be but one perpendicular from a given point to a given line. CD; hence the sum of the the given sum ED. Again, the 2ADC... ABC and ACB are equal AB BE; for a similar reason AC three sides of the triangle ABC angle ABC2AEB, and ACB = the given angles. Q. E. D. THEOREM XXXII. If two triangles have two sides of the one equal to two sides of the other, but the included angles unequal, the third sides will be unequal, and the greater will be in that triangle which has the greater included angle. D F G For at the point D make the angle EDG equal to the angle BAC; take DG equal to AC, and join GE. Then will the triangle DEG equal the triangle ABC (th. 1) and EG BC. = But EG EI + IG, and DF < DI + IF (ax. 13). By addition of these inequalities, EG + DF < EI + IF + DI + IG; or, EG + DF < EF + DG. Taking away the equals DF and DG from the members of the last inequality, there remains EG < EF. But EG - BC ... BC < EF. Q. E. D. If the point G fall within instead of without the triangle DEF, we should have DG + GE < DF + EF (th. 14); and, taking away the equals DG and DF, there remains EG EF. If the point G fall on EF. the theorem is evident. = The converse of this proposition is also true, viz., that if two sides of one triangle be equal to two sides of another, and the third sides unequal, the angle opposite the smaller third side will be less than the one opposite the larger. For if the angle were greater, by the above proposition the third side must be greater; and if it were equal, it must, by (th. 1), be equal. But the third side is neither greater nor equal; therefore the angle opposite, being neither greater nor equal than the angle of the other triangle, must be less. THEOREM XXXIII. Every diameter bisects a circle and its circumference. A B Let ACBD be a circle, AB a diameter. Conceive the part ADB to be turned over and applied to the part ACB, it will coincide with it exactly, otherwise there would be points in the one portion of the circumference or the other unequally distant from the center; but this is contrary to the definition (def. 41); hence the two parts are equal (ax. 10). THEOREM XXXIV. D If a line drawn through the center of a circle bisect a chord, it will be perpendicular to the chord; or, if it be perpendicular to the chord, it will bisect both the chord and the arc of the chord. Let AB be any chord in a circle, and CD a line drawn from the center C to the chord. Then, if the chord be bisected in the point D, CD will be perpendicular to AB. Draw the two radii CA, CB. Then the two triangles ACD, BCD, having CA equal to CB, being ra E D B dii of the same circle (def. 41), and CD common, also AD equal to DB (by hyp.); they have all the three sides of the one equal to all the three sides of the other, and so have their angles also equal (th. 5). Hence, then, the angle ADC being equal to the angle BDC, these angles are right angles, and the line ČD is perpendicular to AB (def. 12). Again, if CD be perpendicular to AB, then will the chord AB be bisected at the point D, or have AD equal to DB; and the arc AEB bisected in the point E, or have AE equal EB. For, the two triangles ACD, BCD being right-angled at D, and having two sides of the one equal to the same in the other, viz., AC = CB and CD common, are equal (th. 26, cor. 2), .. AD = BD. Also, since the angle ACE is equal to the angle BCE, the arc AE, which measures the former, is equal to the arc BE, which measures the latter, since equal angles must have equal measures. Scholium. Two conditions determine a line such as that it shall pass through two given points, or that it shall pass through one point and be perpendicular to a given line, or pass through a point and make a given angle with a given line. The line CE in the last diagram fulfills four conditions. It passes through the center C, through the point D, the middle of the chord, through the point E, the middle of the arc, and, finally, is perpendicular to the chord AB. Either two of these involves the other two. Thus, if a line pass through the middle of the chord and be perpendicular to it, it will pass through the middle of the arc and the center of the circle; if it pass through the middle of the arc and center of the circle, it will pass through the middle of the chord and be perpendicular to it; if it pass through the middle of the arc and chord, it will be perpendicular to the latter, and pass through the center of the circle, &c. THEOREM XXXV. Any chords in a circle which are equally distant from the center are equal to each other; or, if they be equal to each other, they will be equally distant from the center. Let AB, CD be any two chords at equal distances from the center G; then will these two chords AB, CD be equal to each other. Draw the two radii GA, GC, and the two perpendiculars GE, GF, which are E A G |