Proof. If the line C F be not parallel to the bases A B and G E, some other line drawn through C must be parallel to A E. Suppose that the line C D could be parallel to A E, and suppose C D to meet the side G F, or the side GF produced in D. Join the points D and E by the right line D E. We should then have two As, A B C and GED, upon equal bases and between the same parallels. It would follow that these As must be equal in area. (Prop. XXXVIII.) Now since the A A B C is equal in area to the A GEF, it would follow that the A GED would also be equal to the A GEF. But this is impossible, because the one is only a part of the other. The supposition which leads to this impossibility must be itself impossible. That is to say, it is impossible that the line C D, which does not coincide in direction with CF, can be parallel to A E. In the same way it may be proved that no other line which does not coincide with CF can be parallel to A E. Therefore, as there must be a line drawn in some direction from C that is parallel to A E, the line CF, which joins the vertices of the As, must be that line. same PROPOSITION XLI. If a parallelogram and a triangle be upon the base and between the same parallels (or have the same altitude), the parallelogram is the double of the triangle. For the construction in this proposition we must be able to draw a straight line from one given point to another. (Post. I.) For the proof of the proposition we must know,1. That if two s be upon the same base and between the same parallels, they are equal. (Prop. XXXVII.) 2. That a parallelogram is bisected by either of its diagonals. (Prop. XXXIV.) 3. That if a magnitude be the double of one of two equal magnitudes, it is also the double of the other (consequence of the 7th axiom). Let ABCD and A BE be a parallelogram and a A upon the same base, and between the same parallels. It has to be shown that the parallelogram is the double of the A. Draw one of the diagonals of the parallelogram, D с E A B as, e.g., D B. The As A B D and A B E are upon the same base and between the same parallels. It follows (from the 37th proposition) that they are equal to each other (in area). But the parallelogram A B C D is the double of the A ABD. Therefore, it is also the double of the A ABE, which is equal to the A ABD. PROPOSITION XLII. To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. For the construction necessary in this proposition we must be able,1. To join two given points by a straight line. (Post. I.) 2. To bisect a given straight line. parallel to a given straight line. (Prop. XXXI.) 4. From a given point in a given straight line to draw a straight line making with the given Z. (Prop. XXIII.) struction we must know,- same parallels (or having the same altitude), are equal. (Prop. XL.) the same base, and between the same parallels, the parallelogram is the double of the A. (Prop. XLI.) 3. That doubles of the same magnitude are equal to each other. (Ax. VII.) C G Let A B C be the given A, and D E F the given rectilineal Z. Η Κ We have to describe a parallelogram which shall be equal in area to the A A B C, and have an Z equal to the given Z,DEF. A Bisect the line A B in the point G. Through the point C draw a straight line, CHK, parallel to the line A B. From the point G draw a straight line, making with the line A B an 2 equal to the < DEF. Let this line meet the line CHK in the point H. Through B draw a straight line parallel to the line GH, and meeting the line C H produced in the point K. GBKH is the parallelogram required. Proof. Join the points C and G by the straight line CG. The parallelogram G B K H and the A G B C are upon the same base, GB, and between the same parallels, therefore the parallelogram G B K H is double the A GBC. But the As A G C and G B C are upon equal bases, A G and G B (for A B was bisected at the point G), and have the same altitude. Therefore they are equal to one another, and each is the half of the A A BC. Or, in other words, the A ABC is double the A GBC. But the parallelogram G B K H is also double the A GBC. Therefore, the parallelogram GBK H is equal to the A ABC. Moreover, the parallelogram G B KH was structed with an equal to the given Z, DEF. Consequently, G B K H is the parallelogram required. It will do equally well if the parallelogram required be constructed on the base A G instead of G B. con If we take any point in the diagonal of a parallelogram (as, e.g., the point E* in the diagonal D B of the parallelogram A B CD), and through it draw lines parallel respectively to the sides of the parallelogram (as HK and FG), it is obvious that the whole parallelogram is divided into four smaller parallelograms. Those two through which the diagonal of the whole parallelogram passes (forming their diagonals) are called parallelograms about the diagonal of the whole parallelogram. The two other parallelograms, which, together with the parallelograms about the diagonal, fill up the whole parallelogram, are called the complements of the parallelograms about the diagonal. * See the diagram in the next proposition. PROPOSITION XLIII. any two E The complements of parallelograms which about the diagonal of a parallelogram are equal to each other. For proving this proposition we must know,- its diagonals. (Prop. XXXIV.) equal. (Ax. II.) 3. That if equals be taken from equals, the remainders are equal. (Ax. III.) Let ABCD be a parallelogram. Let DB be its diagonal, and KEFD and E GBH two parallelograms D about the diagonal D B. It has to be shown that the parallelograms KEGA and EHCFR (which are the complements of the parallellograms about the diagonal) are equal to each other. Proof. The As D C B and DAB are equal to each other, being the two parts into which the parallelogram A B C D is divided by its diagonal D B. Again, the As DFE and D KE are equal to each other, being the parts into which the parallelogram DFEK is divided by its diagonal, DE; and, for a similar reason, the As EHB and EGB are equal to each other. It follows, therefore, that the sum of the As DFE and EHB, is equal to the sum of the As D KE and EGB. And if these equal sums be taken away from the H A G B |