The line FG intersects the parallels E G and A D, therefore the ADF is equal to the EGF. Consequently ADF is a right . 3. To prove that GE is equal to G F. Since E F intersects the parallels E C and G F, the alternates CEF and EFG are equal. But CEF is half a right, therefore EFG is half a right . And since the threes of the ▲ EF G are together equal to two rights, and EGF is a right, and EFG is half a right, it follows that GE F is half a right, and is equal to the GFE. Consequently (I. 6) G E is equal to G F. 4. To prove that D F is equal to D B. In the ▲ BDF, the BDF is a right, DFB is half a right, therefore D B F is half a right ▲, and is equal to D F B. Therefore D F is equal to D B. 5. To prove that the □ on AF is equal to the sum of thes on AD and D B. ADF is a right, therefore in the ▲ ADF the on AF is equal to the sum of the □s on AD and DF. But D F is equal to DB; therefore the □ on DF is equal to the ☐ on D B. Therefore the on A F is equal to the sum of the Os on AD and D B. 6. To prove that the on A F is also equal to twice the on AC, together with twice the☐ on CD. In the ▲ AE F, the AEF is a right; therefore the on AF is equal to the sum of the s on AE and E F. In the AACE the ACE is a right; therefore the on AE is equal to the sum of the A C and C E, and is therefore equal to twice the A C, since A C and CE are equal. s on on Again, in the AEG F, the EGF is a right <, therefore the on EF is equal to the sum of the s on E G and GF, and is therefore equal to twice the on EG, since E G is equal to G F. But E G and C D are equal, because they are opposite sides of the parallelogram C G. twice the on E G is equal to twice the Therefore on CD. It follows therefore that the on A F is equal to twice the ☐ on A C together with twice the ☐ on C D. 7. It has thus been proved that the on AF is equal to the sum of the Os on AD and D B, and also to twice the on A C, together with twice the on CD. It follows therefore that the sum of the ☐s on A D and D B is equal to twice the on A C, together with twice the on CD. It is an excellent mental exercise to master the elaborate proof given by Euclid of the ninth and tenth propositions; but the result might have been arrived at in a much shorter way. The following is one of the proofs that have been given: Let A B be bisected in C, and divided unequally in D. But 2ce rect. BC, CD + on DB on BC + on CD. (II. 7.) Substitute the second side of this equation for the first in the previous equation, and we get,— on AD + on DB☐ on BC + on BC on CD; i.e., on AD + un B C + 2ce on CD. on CD + on DB=2ce For the 10th proposition, let A B be bisected in C and produced to D. A B ☐on AD☐ on AC + ☐ on CD2ce rect. A C, C D. But CBA C. Substitute CB for AC in the previous equation, and we get, on AD☐ on CB+ Add the on DB ☐on D B. on CD + 2ce rect. CB, C D. on DB to both sides, then on A D + ☐ on CB+ on CD + 2ce rect. CB, CD + But 2ce rect. CB, CD + □ on D B = ☐ on CD + ☐ on C B. (II. 7.) Substitute the second side of this equation for the first in the previous equation, and we get,-~ on AD +☐ on DB =☐ on CB on CB+ on CD; i.e., □ on AD + on CB+2ce on CD. +☐ on CD + on D B = 2ee If in Prop. IX. AD and D B be considered as two separate lines, it is clear that A B is their sum, and A C is half their sum. CD is what is left when D B is taken from CB, therefore it is equal to what would be left if DB were taken from A C. Therefore, if AE be cut off from AC equal to DB, ED will be bisected in C. But ED is the difference between AD and AE or D B, therefore CD is half the difference between A D and D B. It follows, therefore, that the 9th proposition may be thus expressed :-" The sum of the squares on two lines is equal to twice the square on half their sum, together with twice the square on half their difference." A E C D B L The very same relation is proved in the 10th proposition. Let AD and DB be regarded as separate lines. Let BA be produced, and A E cut off equal to BD. E A C B D It is then evident that ED is the sum of AD and DB, and consequently, CD is half AB is the difference between AD and D B, their sum. and therefore A C is half their difference. PROPOSITION XI. To divide a given finite right line so that the rectangle contained by the whole line and one segment shall be equal to the square on the other segment. For the construction required we must be able,— 3. To draw a straght line parallel to a given 4. On a given straight line to describe a square. (I. 46.) For the proof we must know,— 1. That if a line be bisected and produced, the rectangle contained by the whole line and the produced part, together with the on half the line, is equal to the on the line made up of the half and the produced part. (II. 6.) 2. In a right-angled ▲, the☐ on the hypotenuse is equal to the sum of the Os on the other two sides. M Let A B be the given line. On A B describe the ABCD. Bisect DA in E, and join E and B. D K с Produce E A, and cut off E F equal to EB; and from A B cut off AG, equal to A F. G is the point of section required. Now complete the square AH by drawing GH parallel to A F, and FH parallel to A G, and produce HG to meet D C in K. DA is bisected in E, and produced to F; consequently (II. 6), the rect. D F, FA (that is, the rectangle D H), together with the on EA, is equal to the on EF. But E F is equal to E B, therefore the on EF is equal to the on E B. Therefore the rect. DH, together with the on E A, is equal to the □ on E B. But EA B is a right-angled ▲, therefore the☐ on EB is equal to the sum of the Os on EA and A B; and consequently the rect. DH, together with the ☐ on E A, is equal to the sum of the □s on E A and A B. Take away the on E A from both these equals, and we have left the rect. D H equal to the square on A B, that is, the square D B. From these take away the rect. D G, which is common to both, and we have left the □ AH, equal to the rect. K B. But A H is the on A G, and K B is the rect. A B, BG, for it is contained by CB and BG, and CB is equal to A B. Therefore the line AB has been divided so that the rect. A B, B G is equal to the □ on A G. A line thus divided is said to be divided in extreme and mean ratio. |