E A C D B 6. That in a right-angled triangle the on the hypotenuse se is equal to the sum of the Os on the other two sides. (I. 47.) Let A B be a straight line, bisected at C, and divided into two unequal parts at D. We have to prove that the Ion AD together with the on D B is equal to twice the con AC together with twice the on C D. From O draw CE at right Zs to Through D draw DF parallel to CE, and meeting E B in the point F. Through F draw FG parallel to A B, meeting CE in G. Join A and F. The first portion of the proof consists in showing with regard to this figure : 1. That A EF is a right Z. CA is equal to CE, therefore the < CEA is equal to the _CA E. (1. 5.) The three xs of the A ACE are together equal to two right Z8 (I. 32) and ACE is a right Z, therefore CAE and CE A are together equal to one right 2: Consequently, as the <8 CAE and CEA are equal to each other, CE A is half a right Z. In a similar way it be shown that C E B is half a right L. It follows, therefore, that A E B (that is, A EF) is a right 2 2. To prove that ADF is a right : The parallels EC and FD are intersected by CD; therefore the two Zs GCD and CDF are together equal to two right Zs. GCD is a right 2, therefore FDC is a right L 3. To prove that GE is equal to GF: The straight line EC intersects the parallel straight lines GF and CD, therefore the EGF is equal to the < GCD. (I. 29.) But G C D is a right Z, therefore EGF is a right Z. In the AEG F the three xs are together equal to two right Z8; EGF is a right Z, and G EF is half a right Z; consequently, the < GFE is half a right Z, and is equal to the < GEF. It follows, therefore (I. 6), that GE is equal to GF. 4. To prove that D F is equal to DB: The three xs of the AECB are together equal to two right Zs; ECB is a right Z, and C E B is half a right Z; therefore CBE (that is, DBF) is half a right 2. The straight line C B intersects the parallel lines FD and GC; therefore the <FDB is equal to the < GCB. But G C B is a right Z, therefore FDB is a right Z. The three Zs of the A FDB are together equal to two right Zs. (I. 32.) FDB is a right Z, and DBF is half a right Z; consequently, the DFB is half a right Z, and is equal to the < DBF. Because the < DFB is equal to the < DBF, DF is equal to D B. [The remainder of the proof depends upon the fact that AF is the hypotenuse of two right-angled As, namely, AEF and ADF; so that the sum of the Os on A D and :D F admits of being compared with the sum of the s on A E and EF. If this point is clearly kept in mind, the course of the proof will be easily remembered. It will be proved (5) that the on AF is equal to the sum of the Os on A D and D F, or AD and D B; (6) that the on AF being equal to the sum of the Os on AE and E F, is also equal to twice the on A C, together with twice the o on CD; whence it follows (7) that the sum of the Os on A D and D B is equal to twice the con AC together with twice the on CD.] 5. To prove that the con A F is equal to the sum of the Os on A D and D B: ADF is a right Z, therefore the on A F is equal to the sum of the Os on A D and D F. But D F is equal to D B, therefore the on D F is equal to the con D B. Consequently, the sum of the Os on A D and D F is equal to the sum of the Os on A D and DB; and therefore the con A F is equal to the sum of the Os on A D and D B. 6. To prove that the con AF is equal to twice the o on A C, together with twice the con CD: The _ A EF is a right Z, therefore in the A AEF the a on AF is equal to the sum of the Os on A E and EF. (I. 47.) But in the A ACE the < ACE is a right Z, therefore the con A E is equal to the sum of the Os on A C and C E. But AC is equal to CE (by construction), therefore the oon AC is equal to the on CE. Consequently, the sum of the Os on A C and C E is equal to twice the on A C. Therefore the on AE is equal to twice the o on AC. Again, in the A EGF the < EGF is a right Z, therefore the on EF is equal to the sum of the Os on EG and GF, and therefore to twice the a on G.F, because G F is equal to G E. But G F and C D are equal, being opposite sides of the parallelogram GD (I. 34); therefore twice the oon GF is equal to twice the con CD; and, consequently, the on EF is equal to twice the a on CD. It follows, therefore, that the con AF is equal to twice the on A C, together with twice the on CD. 7. It has been proved that the on A F is equal to the sum of the Os on A D and D B, and also to twice the on A C together with twice tbe on CD. It follows, therefore, that the sum of the Os on A D and D B is equal to twice the on A C together with twice the on CD. PROPOSITION X. If a straight line be bisected and produced to any point, the square on the whole line thus formed together with the square on the produced part, is equal to twice the square on half the line together with twice the square on the line made up of the half and the part produced. The propositions required for the construction and proof in this proposition are the same as in the last. Let A B be a right line bisected in C, and produced to D. We have to prove that the con A D together with V the oon DB, is equal to twice the on A C, together with twice the on CD. From C draw CE perpendicular to A B, and cut off CE equal to AC or C B. Join A and E, and B and E. G B A А Through E draw EG parallel to A B, and through D draw D G parallel to EC, meeting E G in G. Because EC and G D are parallel and EG intersects them, the Zs CE G and EG D are together equal to two right Zs. (I. 29.) Therefore the 8 B E G and EG D are together less than two right Zs; and consequently the lines E B and G D will meet if produced. Produce them, and let them meet in the point F. Join A and F. The proof of this proposition is step for step, and almost word for word, the same as that of the last. It is proved, 1. That the _ A EF is a right Z. 5. That the on A F is equal to the sum of the Os on A D and D B. 6. That the on AF is equal to twice the c on A C, together with twice the on CD. 7. Whence it follows, lastly, that the sum of the s on AD and D B is equal to twice the on AC together with twice the on CD. 1. To prove that A EF is a right 2: in the A A CE, AC and CE are equal, therefore the Zs CAE and C E A are equal. And because the three xs of the A ACE are together equal to two right <s, and ACE is a right Z, it follows that CA E and CE A are together equal to a right Z, and therefore C E A is half a right Z. In a similar manner it may be proved that the < CEB is half a right Z. Consequently, A EF is a right Z. In the parallelogram C G the opposite s ECD and E G D are equal (I. 34); therefore E G D is a right 2. |