PROBLEM 3. To bisect a given straight line. Let AB be the given straight line. Construction. With centre A and any radius greater than half AB describe a circle, and with centre B and the same radius describe a circle intersecting the former in two points C and D. Join CD cutting AB in O. Then O will be the point of bisection. Proof. For by the last Problem CD is so drawn as to bisect the angle ACB; And then in the triangles ACO, BCO we have AC= BC, CO common, and the included angles ACO, BCO equal: And therefore by Theorem 16, the base AO = the base BO: that is, O is the point of bisection of AB. PROBLEM 4. To draw a perpendicular to a given straight line from a given point without it. F Let BC be the given straight line, A the given point. Construction. With centre A describe a circle with any sufficient radius to cut BC in two points D, E. Bisect DE in F (Prob. 3). Join AF. Then AF shall be perpendicular to BC. Proof. For if AD, AE be joined, it is clear that the triangles AFD, AFE have the three sides of the one respectively equal to the three sides of the other. Therefore the angle AFD=the corresponding angle AFE (Th. 16), and therefore AF is perpendicular to BC (Def. 10). NOTE. This construction also is usually effected in practice by means of the square. A large class of problems consists of those in which a triangle has to be constructed under certain conditions. We shall take the most important cases, which are also the simplest. PROBLEM 5. To construct a triangle, having given the lengths of the three sides. Let the three given lengths be the lines A, B, C. Construction. Draw a line PQ equal to one of them A. With centre P and radius equal to B describe a circle; and with centre and radius equal to C describe a circle. Let these circles intersect in R. Join RP, RQ. RPQ is the triangle required. Proof. For its three sides are by the construction equal to the three given lines. Limitation. It is necessary that any two of the lines A, B, C should be together greater than the third. For if B and C were together less than A, the circles in the figure would obviously not meet: and if they were together equal to A, the point R would be on PQ, and the triangle would be come a straight line. Similarly if B were greater than A+ C or C greater than A+B, the circles would not intersect. This limitation might be anticipated from the theorem before proved, that any two sides of a triangle are together greater than the third side. COR. I. Hence it is possible with the ruler and compasses alone to effect what is performed in practice generally by means of the sector or protractor, viz. to make an angle equal to a given angle. Join any two points B, C in its sides. Construct a triangle PQR having its three sides PQ, QR, RP respectively equal to AB, BC, CA. Then (by Theorem 18) the angle P is equal to the angle A. COR. 2. An angle equal to a given angle may be made at any point, and such that one of the lines containing it may be any given line through the point. Thus if P were the given point, PS the given line, PQ must be taken equal to AB, and the rest of the construction is the same as before. COR. 3. Hence through any point a straight line can be drawn parallel to a given straight line. Let A be the given point, BC the given line. Draw any line AD to meet BC, and make the angle DAE equal to the alternate angle ADC. Then by Theorem 5, Cor. 2, AE is parallel to BC. PROBLEM 6. To construct a triangle, having given two angles and a side adjacent to both. R Let A, B be the two angles, C the given side. Take a line PQ= C. At the points P, Q make angles equal respectively to A and B (Prob. 5, Cor. 1). Let the lines which contain these angles meet in R. Then RPQ is the triangle required. Limitation. The two given angles must be together less than two right angles, or the lines PQ, QR would not meet. This follows also from the theorem that the three interior angles of a triangle are together equal to two right angles. PROBLEM 7. Having given two angles and a side opposite to one of them, to construct the triangle. Let A and B be the given angles, CD the given side which is to be opposite to A. B = K Construction. Draw an indefinite straight line EF. At any point G in it make the angles FGH=A, and HGK B (Prob. 5), then KGE will equal the third angle of the triangle, since the sum of the three angles of a triangle is equal to two right angles (Th. 7). At C and D make angles equal to HGK and KGE, and let their sides meet in O; then OCD is the triangle required. Proof. For the angle O is the supplement of the angles OCD + ODC, and must therefore be equal to HGF, that is, to A. Limitation.-As before the two given angles must be together less than two right angles. |