son G B 6. That in a right-angled triangle the on the hypotenuse is equal to the sum of the 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 on AD together with the a on D B is equal to twice the con AC together with twice the on CD. From C 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. 2. That A D F is a right Z. 3. That G E is equal to GF. 4. That D F is equal to D B. 1. To prove that A EF is a right : CA is equal to CE, therefore the < CE A is equal to the < CA E. (I. 5.) The three xs of the A ACE are together equal to two right Zs (I. 32) and ACE is a right 2, therefore CAE and CE A are together equal to one right : Consequently, as the <s CAE and CEA are equal to each other, CE A is half a right Z. In a similar way it may be shown that C E B is half a right Z. It follows, therefore, that A E B (that is, A EF) is a right 2 2. To prove that ADF is a right 2: The parallels EC and FD are intersected by CD; therefore the two Zs GCD and CDF are together equal to two right Z8. GOD is a right Z, therefore FDC is a right 2. 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 Zs are together equal to two right Z8; EGF is a right Z, and GEF 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 Zs of the AECB are together equal to two right <s; ECB is a right Z, and C E B is hali a right <; therefore CBE (that is, DBF) is half a right Zi The straight line CB intersects the parallel lines FD and G C; therefore the < FDB is equal to the ZGOB. But GCB is a right Z, therefore FDB is a right Z. The three Zs of the AFDB are together equal to two right Zs. (I. 32.) FDB is a right Z, and DBF is half a right Zi 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 Us 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 (6) that the on AF is equal to the sum of the 8 on AD and D F, or AD and D B; (6) that the on AF being equal to the sum of the Os on AE and EF, 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 8 on AD and D B is equal to twice the on AC together with twice the on CD.] 3. To prove that the con AF is equal to the sum of the Os on A D and DB: ADF is a right Z, therefore the on AF is equal to the sum of the Os on A D and DF. But D F is equal to D B, therefore the on D F is equal to the O on D B. Consequently, the sum of the Us on A D and D F is equal to the sum of the Os on A D and D B; and therefore the on AF is equal to the sum of the Os on A D and D B. 6. To prove that the on AF is equal to twice the o on A C, together with twice the on CD: The ZA EF is a right Z, therefore in the A AEF the on A F 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 o on A E is equal to the sum of the Os on AC and C E. But A C is equal to CE (by constriction), therefore the o on AC is equal to the on CE. Consequently, the sum of the Os on AC and CE is equal to twice the on A C. Therefore the con A E is equal to twice the on AC. Again, in the A EGF the < EGF is a right 2, therefore the on E F is equal to the sum of the Os on EG and GF, and therefore to twice the on GF, because G F is equal to G E. But G F and C D are equal, being opposite sides of the parallelogram G D (I. 34); therefore twice the O on GF is equal to twice the con CD; and, consequently, the on EF is equal to twice the o on OD. It follows, therefore, that the oon A F is equal to twice the O on A C, together with twice the con 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 AC together with twice the con CD. It follows, therefore, that the sum of the Os on A D and D B is equal to twice the con 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 G bisected in C, and produced to D. We have to prove that the on AD together with A D the con DB, is equal to twice the a on A C, together with twice the con CD. From O draw CE perpendicular to A B, and cut off CE equal to AC or C B. Join A and E, and B and E. B в Through E draw E G parallel to A B, and through D draw D G parallel to E C, meeting E G in G. Because EC and G D are parallel and EG intersects them, the Z8 CE G and E G D are together equal to two right Zs. (I. 29.) Therefore the s B E G and EGD 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 Z A EF is a right Z. 5. That the on AF is equal to the sum of the Os on AD and D B. 6. That the on AF is equal to twice the on AC, together with twice the on CD.- 7. Whence it follows, lastly, that the sum of the 8 on AD and D B is equal to twice the on A Ć together with twice the on CD. 1. To prove that A EF is a right 2: in the A ACE, AC and C E 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 CAE and CE A are together equal to a right 2, and therefore CEA is half a right 2. In a similar manner it may be proved that the < CEB is half a right Z. Consequently, A EF is a right 2. In the parallelogram C G the opposite Z8 ECD and E G D are equal (I. 34); therefore E G D is a right Z. |