PROP. V. THEOREM. If a straight line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line. Let AB be a straight line divided into two equal parts in C, and into two unequal parts in D. Then it is to be proved that The rectangle AD, DB together} = The square on CB. with the square on CD CONSTRUCTION.-1. On CB describe the square CEFB (I. 46), and join BE. 2. Through D draw DHG parallel to CE or BF (I. 31), cutting BE in H. 3. Through H draw KLHM parallel to AB or EF, cutting CE in L and BF in M. 4. Through A draw AK parallel to CL or BM. PROOF.-Because the comp. CH= the comp. HF (I. 43), Therefore CH and DM = HF and DM (ax. 2), the whole DF. AL (I. 36), therefore AL DF (ax. 1), and therefore AL and CH = DF and CH (ax. 2.), i.e. the rect. AH = the gnomon CMG. But because AH is the rect. AD, DH, and DH = DB (II. 4, Cor.), therefore AH = the rect. AD, DB, and therefore the gnomon CMG = the rect. AD, DB (ax. 1); therefore also the gnomon CMG together with LG, i.e. the whole figure CEFB = { the rect. AD, DB, together with LG; But because LG is the square on LH (II.4, Cor.) and LH=CD (I. 34), therefore the whole figure CEFB-(the rect. AD, DB, together with the square on CD. But the whole figure CEFB = the square on CB (cons.), Therefore, it is proved, as required, that The rectangle AD, DB, together with the square on Wherefore, If a straight line be divided, &c. Exercises. Q. E. D. 1. Prove that as stated in Prop. IV. Proof 3, the figure HF is a square on the side HG. 2. Prove that in Prop. V. the rectangle AD, DB, together with the square on CD = the square on AC. PROP. VI. THEOREM. If a straight line be bisected, and produced to any point, the rectangle contained by the whole line thus produced, and the part of it produced, together with the square on half the line bisected, is equal to the square on the straight line which is made up of the half and the part produced. Let AB be a straight line bisected in C, and produced to ny point D. CONSTRUCTION.-1. On CD describe the square CEFD (I. 46), and join DE. 2. Through B draw BG parallel to CE or DF (I. 31), cutting DE in H. 3. Through H draw KM parallel to AD or EF, cutting CE in L and DF in M. 4. Through A draw AK parallel to CL or DM. But because AM is the rect. AD, DM, and DM = DB (II. 4, Cor. ), therefore AM rect. AD, DB, and therefore the gnomon CMG = the rect. AD, DB (ax. 1), therefore also the gnomon the rect. AD, DB to CMG together with LG, = gether with LG. i.e. the whole figure CEFD And because LG is the square on LH (II. 4, Cor.), and LH = CB (I. 34), therefore the whole figure CEFD = the rect. AD, DB, to gether with the square on CB. But the whole figure CEFD = the square on CD (cons.). Therefore, it is proved, as required, that The rectangle AD, DB together with the square on BC the square on CD. Wherefore, If a straight line be bisected, &c. Q. E. D. Exercise. Prove that if in the above Proposition we take AB as bisected in C, and produced beyond A to D, then we should have, The rectangle BD, AD together with the square on AC the square on CD. PROP. VII. THEOREM. If a straight line be divided into any two parts, the squares on the whole line, and on one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square on the other part. C. Let AB be a straight line divided into any two parts in CONSTRUCTION.-1. On AB describe the square ADEB (I. 46), and join BD. 2. Through C draw CF parallel to AD or BE (I. 31), cutting BD in G. 3. Through G draw HK parallel to AB or DE. PROOF. Because AG = GE (I. 43), therefore AG and CK = GE and CK (ax. 2), i.e. AK = CE; and therefore AK and CE = twice AK. But AK and CE = the gnomon AKF and CK, therefore the gnomon AKF and CK = twice AK (ax. 1). |