Again, the ext. angle DQB the int. opp. angle CPB; I. 29. ... the angle DQB is a rt. angle. And the angle QBD is half a rt. angle; Proved. .. the remaining angle QDB is half a rt. angle; 1. 32. .. the angle QBD = the angle QDB; ... QD=QB. = Now the sq. on AP the sq. on PC; for AP PC. Constr. And since the angle APC is a rt. angle, .. the sq. on AC = the sum of the sqq. on AP, PC; I. 47. ..the sq. on AC is twice the sq. on AP. Similarly, the sq. on CD is twice the sq. on ED, that is, twice the sq. on the opp. side PQ. Now the sqq. on AQ, QB = the sqq. on AQ, QD That is, = the sq. on AD, for AQD = angle; I. 34. Proved. is a rt. I. 47. = the sum of the sqq. on AC, CD, for ACD is a rt. angle; I. 47. =twice the sq. on AP with twice the sq. on PQ. Proved. the sum of the sqq. on AQ, QB = twice the sum of the sqq. on AP, PQ. Q.E.D. [NOTE. For alternative proofs of this proposition, see page 148.] PROPOSITION 10. THEOREM. [EUCLID'S PROOF.] If a straight line is bisected and produced to any point, the sum of the squares on the whole line thus produced, and on the part produced, is twice the sum of the squares on half the line bisected and on the line made up of the half and the part produced. Let the st. line AB be bisected at P, and produced to Q. Then shall the sum of the sqq. on AQ, QB be twice the sum of the sqq. on AP, PQ. Construction. At P draw PC at right angles to AB; I. 11. and make PC equal to PA or PB. Join AC, BC. I. 3. Through Q draw QD par to PC, to meet CB produced in D ; I. 31. and through D draw DE par' to AB, to meet CP produced in E. Proof. Join AD. Then since PA = PC, the angle PAC = the angle PCA. Constr. And since, in the triangle APC, the angle APC is a rt. angle, the sum of the angles PAC, PCA is a rt. angle. I. 32. Hence each of the angles PAC, PCA is half a rt. angle. So also, each of the angles PBC, PCB is half a rt. angle. .. the whole angle ACB is a rt. angle. Again, the ext. angle CPB the int. opp. angle CED: I. 29. ..the angle CED is a rt. angle: and the angle ECD is half a rt. angle ; Proved. .. the remaining angle EDC is half a rt. angle. I. 32. .. the angle ECD = the angle EDC; I. 6. Again, the angle DQB= the alt. angle CPB; I. 29. Also the angle QBD=the vert. opp. angle CBP: 1. 15. that is, the angle QBD is half a rt. angle. .. the remaining angle QDB is half a rt. angle: I. 32. ... the angle QBD = the angle QDB; ... QB=QD. Now the sq. on AP = the sq. on PC; for AP = PC. == I. 6. Constr. .. the sq. on AC = the sum of the sqq. on AP, PC; I. 47. .. the sq. on AC is twice the sq. on AP. Similarly, the sq. on CD is twice the sq. on ED, that is, twice the sq. on the opp. side PQ. Now the sqq. on AQ, QB = the sqq. on AQ, QD That is, I. 34. Proved. =the sq. on AD, for AQD is a rt. angle; = I. 47. the sum of the sqq. on AC, CD, for ACD is a rt. angle; I. 47. =twice the sq. on AP with twice the sq. on PQ. Proved. the sum of the sqq. on AQ, QB is twice the sum of the sqq. on AP, PQ. Q.E.D. CORRESPONDING ALGEBRAICAL FORMULA. The result of this proposition may be written AQ2+QB2=2(AP2+ PQ2). Let AB=2a; and PQ=b; Hence we have then AP and PB each=a. Also AQ=a+b; and QB=b-a. (a+b)2+(b-a)2=2(a2+b2). [NOTE. For alternative proofs of this proposition, see page 149.] PROPOSITION 9. [ALTERNATIVE PROOF.] If a straight line is divided equally and also unequally, the sum of the squares on the two unequal parts is twice the sum of the squares on half the line and on the line between the points of section. P Let the straight line AB be divided equally at P and unequally at Q. Then shall the sum of the sqq. on AQ, QB be twice the sum of the sqq. on AP, PQ. Proof. The sq. on AQ= the sum of the sqq. on AP, PQ with twice the rect. AP, PQ II. 4. the sum of the sqq. on AP, PQ with twice the rect. PB, PQ; for PBAP. To each of these equals add the sq. on QB. Then the sqq. on AQ, QB = the sum of the sqq. on AP, PQ with twice the rect. PB, PQ and the sq. on QB. But twice the rect. PB, PQ and the sq. on QB the II. 7. = the sum of the sqq. on PB, PQ. sqq. on AQ, QB = the sum of the sqq. on AP, PQ with the sum of the sqq. on PB, PQ = twice the sum of the sqq. on AP, PQ. Q.E.D. NOTE. The following concise proof, obtained from II. 4 and II. 5, is useful as an exercise, but it is hardly admissible as a formal demonstration owing to its algebraical use of the negative sign. PROPOSITION 10. [ALTERNATIVE PROOF.] If a straight line is bisected and produced to any point, the sum of the squares on the whole line thus produced and on the part produced, is twice the sum of the squares on half the line bisected and on the line made up of the half and the part produced. Let the straight line AB be bisected at P, and produced to Q. Then shall the sum of the sqq. on AQ, QB be twice the sum of the sqq. on AP, PQ. Proof. The sq. on AQ=the sum of the sqq. on AP, PQ with twice the rect. AP, PQ = II. 4. -the sum of the sqq. on AP, PQ with twice the rect. PB, PQ; for PB AP. = To each of these equals add the sq. on QB. Then the sqq. on AQ, QB = the sum of the sqq. on AP, PQ with twice the rect. PB, PQ and the sq. on QB. But twice the rect. PB, PQ and the sq. on QB = the sum of the sqq. on PB, PQ. II. 7. .. the sqq. on AQ, QB = the sum of the sqq. on AP, PQ with the sum of the sqq. on PB, PQ = twice the sum of the sqq. on AP, PQ. Q.E.D. NOTE. Another proof of this proposition, based on II. 7 and II. 6, is indicated by the following steps: We have AQ2+QB2=2AQ. QB+AB2 =2AQ. QB+4PB2 II. 7. II. 4, Cor. 2. II. 6. |