PROBLEM 2. To draw a perpendicular to a given straight line from a given point given in it. Construction. With centre C and any radius describe a circle to cut the straight line in two points D, E, so that CD CE. With centre D, and any greater than DC, describe a circle, and with centre E and the same radius describe a circle, cutting the former in F. Join FC; it is required to prove that FC is perpendicular to AB. Proof. In the triangles DCF, ECF, and the base DF= the base EF, therefore the angle DCF= the angle ECF, and therefore DCF and ECF are right angles*. (Constr.) (Constr.) (Th. 15.) (Def. 14.) NOTE.-This construction is usually effected in practice by means of the square. It may be observed that this problem is only a special case of Prob. 1, the given angle being a straight angle. PROBLEM 3. To draw a perpendicular to a given straight line from a given point outside it. Let BC be the given straight line, A the given point. * Euclid, I. 11. Construction. With centre A describe a circle with any sufficient radius to cut BC in two points D, E. B Bisect the angle DAE by the line AF. (Prob. 1.) Proof. In the triangles AFD, AFE, because AD=AE, (Constr.) and AF is common, and the contained angle DAF- the contained angle EAF; (Constr.) therefore the angle AFD = the angle AFE; therefore AF is perpendicular to DE*. (Th. 5.) NOTE. This construction also is usually effected in practice by means of the square. PROBLEM 4. 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 will be the point of bisection. Join AD, DB. = Proof. Because AC CB; and CD is common to the two triangles ACB, DCB; and the base AD is equal to the base DB; therefore the angle ACD = the angle BCD; therefore in the two triangles ACO, BCO, AC= BC, we have CO common, (Constr.) and the included angles ACO, BCO equal; therefore the base AO the base BO, = or the line AB is bisected in O. (Th. 5.) 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. to B describe a circle; and ABC with centre Q and radius equal to C describe a circle. Let these circles intersect in R. Join RP, RQ. RPQ is the triangle required. Proof. For RPQ has its three sides respectively equal to A, B and C*. Euclid, I. 22. 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 become 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, and is in fact its contrapositive. PROBLEM 6. At a given point in a given straight line to make an angle equal to a given angle. Let BAC be the given angle, P the given point in the line PQ. Constr. Join any two points B, C in the arms of the given angle. Construct a triangle PQR having its three sides PQ, QR, RP respectively equal to AB, BC, CA. PROBLEM 7. To draw through any point a straight line parallel to a given straight line. Constr. 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. (Prob. 6.) Proof. Because the alternate angles EAD, ADC are equal, therefore AE is parallel to DE*. PROBLEM 8. (Constr.) (Th. 21.) To construct a triangle, having given two angles and a side adjacent to both. P Let A, B be the two angles, C the given side. Take a line PQ= C. At the points P, Q make angles with PQ equal respectively to A and B. (Prob. 6.) Let the lines which contain these angles meet in R. Then RPQ is the triangle required. Proof. For it has PQC, and the angles P and Q respectively equal to A and B. * Euclid, 1. 31. |