Let ABCD, EBCF be parallelograms on the same base BC, and between the same parallels AF, BC. They shall be equivalent. Proof. For in the triangles BAE, CDF the angle A = the angle D, and the angle E= the angle F, by parallelism (Th. 4), and the side AB = the corresponding side DC, and therefore the triangles are equal in area, by Theorem 15. But if the triangle CDF is taken away from the trapezium ABCF, the parallelogram ABCD remains: and if the triangle BAE is taken away from the same trapezium the parallelogram EBCF remains. Therefore these parallelograms are equivalent. Remark. This theorem is the fundamental theorem of equivalence of areas. COR. 1. A parallelogram is therefore equivalent to the rectangle on the same base with the same altitude, and its area is therefore determinate when its base and altitude are known. COR. 2. Parallelograms on equal bases and between the same parallels are equivalent. Proof. For if the base AB= the base EF, the parallelogram AEFG can be conceived as superposed on the parallelogram ABCD, EF coinciding with AB, so that they should have the same base, and be between the same parallels, and therefore they are equivalent. COR. 3. Parallelograms on equal bases and of the same altitude are equivalent. For similarly they may be conceived as placed on one another so as to have the same base, since the bases are equal, and so as to be between the same parallels, since the altitudes are equal; and therefore they are equivalent. THEOREM 23. Triangles on the same base and between the same parallels will be equivalent. Let ABC, DBC be triangles on the same base BC and between the same parallels, they shall be equivalent. B Proof. For if BE be drawn from B parallel to AC, and CF be drawn from C parallel to BD, to meet AD produced in E and F, Then ACBE, DBCF are parallelograms, and are equivalent to one another (by Theorem 22). But the triangles ABC, DBC are respectively the halves of the parallelograms ACBE, DBCF (Th. 21). Therefore the triangles are equivalent. COR. I. Triangles on equal bases and between the same parallels are equivalent. COR. 2. Triangles on equal bases and of the same altitude are equivalent. Hence a triangle is of determinate area when its base and altitude are given. COR. 3. If a parallelogram and a triangle have equal bases and altitudes, the parallelogram is double of the triangle. Proof. For the parallelogram ACBE is double of the triangle ABC, and therefore of any other triangle DBC which is on the same base, and between the same parallels. THEOREM 24. Conversely. Equivalent triangles on equal bases will be of equal altitude. For let the triangles ABC, ABD be equivalent triangles on equal bases, and let them be conceived as placed on the same base AB, and on the same side of it. Then shall their altitudes be A equal, that is, CD will be parallel to AB. E D Proof. For if CD were not parallel to AB, suppose that some other line through C, as CE, meeting AD in E, were parallel to AB. Then the triangles CAB, EAB would be equivalent by Theorem 23. But we know that CAB, DAB are equivalent, therefore EAB and DAB would be equivalent, which is absurd. COR. Hence equivalent triangles on the same base, and on the same side of it, must be between the same parallels. Def. 37. A rectangle is determined when two of its adjacent sides are known. It is then said to be contained by its two adjacent sides, or by lines equal to them, Def. 38. A line AB is said to be internally divided in P when Plies between A and B: and it is said to be externally divided in P when P lies in AB produced: and AP, BP are called in both cases the segments of AB. THEOREM 25. If a straight line is divided internally in any point, the square on the line will be equal to the squares on the two segments together with twice the rectangle contained by the segments. Let AB be divided internally Through P draw PLF parallel to AD, meeting DE in F: cut off PL= PB. Through L draw HLM parallel to AB, to meet DA and EB in M, H. Then it may be seen that the figures AL, PH, LE, MF are parallelograms by construction; and it is easily shewn that PH, MF are the squares on PB, AP respectively; and that AL, LE are each of them the rectangle contained by AP, PB. Hence, since ADEB is made up of these four figures, it follows that the square on AB is equal to the squares on AP, PB and twice the rectangle contained by AP, PB. THEOREM 26. If a straight line be divided externally in any point, the square on the line is equal to the squares on the two segments diminished by twice the rectangle contained by the segments. Then as before it is evident that MF is the square on AP; and MP or HF, the rectangle contained by AP and BP; and HP the square on BP. And AE is less than MF+HP by MP+HF; that is, the square on AB is equal to the squares of AP, PB diminished by twice the rectangle contained by AP, PB. THEOREM 27. The rectangle contained by the sum and difference of any two lines is equivalent to the difference of the squares on those lines. contained by AC, AD, that is, by the sum and difference of AB, BC. |