2. That parallelograms upon the same base and between the same parallels are equal. (Prop. XXXV.) 3. That the halves of equals are equal. (Ax. VIII.) Let ABC and ABD be two As upon the same base, A B, and between the same parallels, AB and It has to be shown that they are equal in EF. area. From the point A draw a line A E, parallel to the side B C of the AA BC, and meeting the line E F in the point E. From the point B draw a line, BF, parallel to the side A D, of the A ABD, and meeting the line E F in the point F.* We thus get two parallelograms, A BCE and • It will answer the purpose just as well, if from the point A a line be drawn parallel to the line BD, and from the point B a line parallel to the line AC. All that we have to attend to is, that the line drawn from the point A be parallel to a side of one of the As, and that the line drawn from the point B be a parallel to a side of the other A. It would also do equally well if from the point A two lines were drawn, one parallel to a side of one A, and the other parallel to a side of the other. Or, in like manner, two lines might be drawn from B, one parallel to a side of one A, the other parallel to a side of the other. The sides to which the lines are to be drawn parallel are of course those which do not terminate in the point from which the said lines are drawn. All these varieties of construction effect the same object, viz., to get two parallelograms upon the same base, of which one is the double of one of the 28 and the other of the other. ABFD, upon the same base, A B, and between the same parallels, A B and EF. These parallelograms, (according to the 35th proposition,) will be equal. But the parallelogram ABCE is the double of the A ABC, being divided into two equal parts, of which the A ABC is one, by its diagonal AC; and the parallelogram A BFD is the double of the A ABD, being divided into two equal parts, of which the A A B D is one, by its diagonal, BD. (Prop. XXXIV.) Consequently, as the halves of equals are equal, the A ABC is equal to the A ABD.* PROPOSITION XXXVIII. Triangles upon equal bases and between the same parallels are equal. For the construction in this proposition we must be able to draw a straight line from a given point so as to be parallel to a given straight line. (Prop. XXXI.) To prove the proposition by the aid of the con struction we must know,1. That parallelograms upon equal bases and between the same parallels are equal. (Prop. XXXVI.) 2. That a parallelogram is divided into two equal parts by either of its diagonals. (Prop. XXXV.) 3. That the halves of equals are equal. (Ax. VIII.) * That is, in area. They have not been shown to be equal in any other respects. Beginners often confound the being equal in area with the being equal in every respect. N A BE F Let A B D and EFH be two As upon equal bases A B and EF, and between D C на the same parallels, AF and DG. It has to be shown that they are equal in area. From either extremity of A B (as B) draw a line (BC) parallel to that side of the A ABD which does not terminate in B, and meeting the line D G in the point C. From either extremity of EF (as F) draw a line (FG) parallel to that side of the A EFH which does not terminate in F. We thus get two parallelograms, A B C D and EFGH, which, being upon equal bases and between the same parallels, are equal. (Prop. XXXVI.) Now the parallelogram ABCD is the double of the A ABD, being divided into two equal parts, of which the A ABD is one, by its diagonal B D. Also the parallelogram E F G H is the double of the A EFH, being divided into two equal parts, of which the A EFH is one, by its diagonal FH. Consequently, since the halves of equals are equal, the A ABD is equal (in area) to the A EFH. PROPOSITION XXXIX. If two triangles which are equal in area be upon the same base and upon the same side of it, the right line which joins their vertices will be parallel to the base upon which they stand.* * This proposition is generally stated thus :-"Equal As upon the same base, and upon the same side of it, are between the same parallels.” The meaning of this is just what is stated above, but is less obvious to the beginner. For the construction in this proposition we must be able to join two given points by a straight line. For proving the proposition by the aid of the con struction, we must know, That As upon the same base and between the same parallels are equal. A B A B The proof of this proposition is of the indirect kind. Let A B C and A B D be two As of equal area, upon the same base, A B, and upon the same side of it. It has to be proved that the line CD, which joins their vertices, is parallel to the base, A B. Proof. If the line C D be not parallel to the base A B, some other line, drawn through C, must be parallel to the base. Suppose the line CE could be parallel to the base, and suppose C E to meet the side A D, or the side AD produced, in the point E. Join the points E and B by the straight line E B. We should then have two As, A CB and A E B, upon the same base, A B, and between the same parallels, A B and CE. It would follow, therefore, that these As would be equal to one another. (Prop. XXXV.) But we know, to start with, that the A A BC is equal to the A ABD; therefore the A ABE would be equal to the A ABD. Now this is an impossibility, for the one is only a part of the other; consequently, the supposition that leads to this impossibility must be itself impossible. That is to say, it is impossible that the line CE should be parallel to A B. In a similar manner it may be shown that no line drawn from the point C can be parallel to the base A B, if it does not coincide with the line CD. Therefore, as there must be a line, drawn in some direction from C, that is parallel to A B, the line CD must be that line. PROPOSITION XL. If two equal triangles stand upon equal bases, which are in the same straight line, the straight line joining the vertices of the triangles will be parallel to the bases.* For the proof of this proposition we must be able to join two given points by a straight line; and must know that As upon equal bases and between the same parallels are equal. (Prop. XXXVIII.) Let A B C and G E F be two As of equal area, and standing upon equal bases A B and G E, which are in the same straight line. It has to be shown that the D line C F, which joins their vertices, is parallel to the line of the bases A B and G E. * This proposition 'is usually stated thus :-Equal As upon equal bases in the same straight line are between the same parallels. |