6. That in a right-angled triangle the on the hypotenuse is equal to the sum of the Os on the other two sides. (I. 47.) E Let A B be a straight line, bisected at C, and divided into two unequal parts at D. We have to prove that the on A D together with the on D B is equal to twice the twice the on AC together with on CD. From C draw CE at rights to A B, and cut off C E equal to CA or C B. G D B Through D draw DF parallel to CE, and meeting E B in the point F. Through F draw FG parallel to AB, meeting CE in G. Join A and F. The first portion of the proof consists in showing with regard to this figure: 1. To prove that A E F is a right ▲ :— CA is equal to CE, therefore the CEA is equal to the CAE. (I. 5.) The threes of the ▲ ACE are together equal to two rights (I. 32) and ACE is a right, therefore CAE and CE A are together equal to one right. Consequently, as the s CAE and CEA are equal to each other, CEA is half a right . In a similar way it may be shown that CE B is half a right . It follows, therefore, that A E B (that is, A EF) is a right <. 2. To prove that ADF is a right ▲ :— The parallels EC and FD are intersected by CD; therefore the two s GCD and CDF are together equal to two rights. GCD is a right, therefore FDC is a right Z. 3. To prove that GE is equal to G F :— The straight line EC intersects the parallel straight lines GF and CD, therefore the EGF is equal to the GCD. (I. 29.) But GCD is a right, therefore E G F is a right <. In the EGF the threes are together equal to two rights; EGF is a right, and G E F is half a right; consequently, the GFE is half a right <, and is equal to the GEF. It follows, therefore (I. 4. To prove that DF is The threes of the 6), that G E is equal to G F. equal to DB: ECB are together equal to two rights; ECB is a right, and CEB is half a right ; therefore CBE (that is, DBF) is half a right . The straight line CB intersects the parallel lines FD and GC; therefore the FDB is equal to the GCB. But GCB is a right, therefore F D B is a right ▲. The threes of the ▲ FDB are together equal to two rights. (I. 32.) FD B is a right, and DB F is half a right ; consequently, the DFB is half a right, and is equal to the ▲ D B F. 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 s on AD and DF admits of being compared with the sum of the s on AE 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 s on AD and D F, or AD and DB; (6) that the on A F 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 on CD; whence it follows (7) that the sum of the s on AD and D B is equal to twice the on AC together with twice the on CD.] 5. To prove that the on AF is equal to the sum of thes on AD and D B: : ADF is a right, therefore the on A F is equal to the sum of the s on AD and D F. But D F is equal to D B, therefore the □ on D F is equal to the on D B. Consequently, the sum of thes on AD and D F is equal to the sum of the s on AD and D B; and therefore the☐ on AF is equal to the sum of the s on A D and D B. 6. To prove that the on AF is equal to twice the on A C, together with twice the on CD:The AEF is a right, therefore in the▲ AEF the on AF is equal to the sum of the Os on A E and EF. (I. 47.) : But in the ▲ ACE the ACE is a right, therefore the on A E is equal to the sum of the s on A C and CE. But AC is equal to CE (by construction), therefore the on AC is equal to the on CE. Consequently, the sum of the □s on A C and CE is equal to twice the on A C. Therefore the on AE is equal to twice the on A C. Again, in the AEGF the EGF is a right, therefore the on EF is equal to the sum of the s on EG and G F, and therefore to twice the on G F, because G F is equal to G E. But G F and CD are equal, being opposite sides of the parallelogram G D (I. 34); therefore twice the on GF is equal to twice the on CD; and, consequently, the on EF is equal to twice the on CD. It follows, therefore, that the ☐ on A F 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 □s on AD and D B, and also to twice the on AC together with twice the on CD. It follows, therefore, that the sum of the Os on AD and D B is equal to twice the on A C together with twice the on CD. PROPOSITION X. on If a straight line be bisected and produced to any point, the square on the whole line thus formed together with the square 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. 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. Through E draw EG parallel to A B, and through D draw DG parallel to E C, meeting E G in G. Because EC and G D are parallel and E G intersects them, the s CEG and EGD are together equal to two rights. (I. 29.) Therefore the s B E G and EGD are together less than two right s; 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 AEF is a right . 2. That ADF is a right 3. That GE is equal to G F. 4. That D F is equal to D B. 5. That the AD and D B. 6. That the on A F is equal to the sum of the s on on A F is equal to twice the ☐ on A C, together with twice the on CD. 7. Whence it follows, lastly, that the sum of the ☐s on AD and DB is equal to twice the on AC together with twice the on CD. 1. To prove that A EF is a right : In the ▲ A CE, AC and CE are equal, therefore thes CAE and CE A are equal. And because the threes of the ▲ ACE are together equal to two rights, and ACE is a right, it follows that CAE and CE A are together equal to a right, and therefore CE A is half a right . In a similar manner it may be proved that the CEB is half a right . Consequently, A EF is a right . 2. To prove that A D F is a right ▲: In the parallelogram C G the opposite s E CD and E G D are equal (I. 34); therefore E G D is a right Z. |