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. (Prob. 5.) Proof. In the triangles ABC, PQR, because AB= PQ, (Constr.) AC = PR, (Constr.) BC = QR, (Constr.) therefore the angle A = the angle P*. (Th. 12.) * Euclid, 1. 23. and 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, (Constr.) therefore AE is parallel to DE* (Th. 21.) PROBLEM 8. To construct a triangle, having given two angles and a side adjacent to both. 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. Proof. For it has PQ=C, and the angles P and I respectively equal to A and B. * Euclid, 1. 31. 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 any two interior angles of a triangle are together less than two right angles, and is the contrapositive of that theorem. PROBLEM 9. To construct a triangle, having given two angles and a side opposite to one of them. Let A and B be the given angles, CD the given side which is to be opposite to A. Construction. Draw an indefinite straight line EF. At any point G in it make the angles FGH= A, and HGK=B (Prob. 6), 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. 25). At C and D make angles equal to HGK and KGE, and let their sides meet in 0; then OCD is the triangle required. Proof. For OCD has CD equal to the given line, and the angles C and D equal respectively to the given angles. Limitation. As before, the two given angles must be together less than two right angles. PROBLEM 10. To construct a triangle, having given two sides and the angle between them. Let A, B be the given sides, C the given angle. Construction. Draw an angle D equal to the given angle, and take DE, DF equal to A and B. Join EF. Proof. For the triangle DEF has DE, DF equal to the given lines A and B, and the included angle D equal to the given angle C. Remark. In these problems we have found that one triangle and only one can be constructed to fulfil the conditions given. In other words, that with these data the triangle is determinate. Also we notice that in each case three elements in the triangle are data or given. We have given either the three sides, or two angles and the side adjacent to both, or two angles and a side opposite to one, or two sides and the included angle. And these cases correspond to the theorems proved above of the equality of triangles. For if only one triangle can be constructed so as to have its sides equal to three given lines, it is clear that if two triangles have the three sides of the one equal to the three sides of the other, these triangles must be identical, or be equal in all respects. And a similar remark may be made on the other cases we have con. sidered. But there are cases in which the data may be insufficient to determine the triangle. For example, if only two sides are given, an indefinite number of different triangles may be constructed to have these sides. Or if the three angles are given, their sum being equal to two right angles, an indefinite number of triangles may be constructed to have these three angles. And again it may be impossible to construct the triangle with the given data, as has been already shewn. In some cases moreover the solution is ambiguous, that is, there may be more than one triangle which fulfils the given conditions. The following is an important instance of this, and is usually called the ambiguous case, some consideration of which occurred in Theorem 20. PROBLEM II. To construct a triangle, having given two sides and an angle opposite to one of them. Let A, B be the given sides, C the angle to be opposite to the side B. Take an angle GDH = C, take DE = A, and with centre E and radius= B describe a circle. If I is one of the points in which this circle meets the line DH, by joining EI we obtain a triangle which fulfils the given conditions. But several cases may arise. (1) If B is less than the perpendicular from E on DH, the circle would not meet DH, and the triangle would be impossible. |