Hence AC must meet B'C' either in C' or E. If it meet B'C' in C', the triangles ABC, A'B'C' are congruent; otherwise not. Now if C and C' be both acute or both obtuse angles, the points C and C' will both lie on the same side of the perpendicular, and will coincide. If one of the angles C and C' be acute and the other obtuse, the triangles are not congruent. If one of the angles as C' be a right angle, A'C' will coincide with A'D. In this case AC must also coincide with A'D, since no other line equal to A'D can be drawn for A' to B'C'. Hence if the angles C and C' be both acute or both obtuse, or if either of them be a right angle, the triangles are congruent. COR. 1. If the angle C be not equal to C', it is supplementary to it. COR. 2. If B and B' the angles given equal be opposite to the greater sides, the triangles must be congruent. For in this case C and C' must be both acute. Hence if the angles given equal be right angles the triangles must be congruent. This is a very common and important case, and admits easily of an independent demonstration. For if AB, A'B' be placed so as to coincide, AC, A'C' being equal must be equally distant from the perpendicular. Therefore BC= B'C', and the triangles are congruent. PARALLELOGRAMS. A Parallelogram is a four-sided figure which has its opposite sides parallel. THEOREM XVI. The opposite sides and angles of parallelograms are equal. Let ABCD be a parallelogram. Then shall AB= CD, BC = DA, A B Then the alternate angle ABD = the alternate angle CDB, [1. 8. Cor. 2. and the alternate angle BDA = the alternate angle DBC. Therefore in the triangles ABD, CDB, [I. 8. Cor. 2. ABD=CDB, BDA=DBC, and BD is common. Therefore AB= CD, BC=DA, and A=C. [I. 15. Also the whole angles ABC, CDA, the parts of which are separately equal, are equal. THEOREM XVII. The straight lines which join the extremities of equal and parallel straight lines towards the same parts are themselves equal and parallel. Let AB be equal and parallel to CD. Then shall BC be equal and pa rallel to DA. Join BD. A B D Then the alternate angle ABD = the alternate angle CDB. Hence in the triangles ABD, CDB, AB= CD, BD is common, and ABD = CDB. Therefore DA=BC, and BDA = DBC, · Therefore but BDA, DBC are alternate angles. BC is parallel to DA. [I. 14. [I. 7. Cor. 2. THEOREM XVIII. The exterior angles of any convex polygon are together equal to four right angles, and the interior angles to twice as many right angles as the figure has sides, diminished by four right angles. D' B' ** Let ABCDE be any convex polygon. B Through any point O draw OB', OC', OD', OE', OA', parallel to the sides of the polygon. and Then the angle C'OB' = the exterior angle at B, D'OC' the exterior angle at C. = Similarly the other angles at 0= the other exterior angles at D, E, and A. Therefore all the exterior angles = all the angles at O Again, each interior angle with its adjacent exterior angle makes up two right angles. Therefore all the interior and all the exterior angles are together equal to twice as many right angles as the figure has sides. Hence the interior angles are equal to twice as many right angles as the figure has sides, diminished by four right angles. COR. The exterior angle of any regular polygon of n sides is equal to one nth part of four right angles. Loci. THEOREM XIX. Through the middle point of AB let CD be drawn perpendicular to AB. Then every point on CD is equally distant from A and from B. For if P be any point on CD, and PA, PB be joined, these lines must be equal, being equally remote from the perpendicular. Also no point not on CD can be equally distant from A and B. Let F be any point not on CD. Join FA, FB; let FA cut CD in E; join EB. Hence if we know of the point P that it must be equally distant from A and from B, we shall know that it must lie somewhere on CD. For example, if P be the centre of any circle passing through A and B, P must be equally distant from A and B; therefore it must lie somewhere on CD. In such a case CD is called the locus of P. THEOREM XX. Let AB, AC be two straight lines which meet in A, and let AD bisect the angle BAC. |