SUMMARY OF II. 5, 6, AND 8. GENERAL ENUNCIATION. The rectangle contained by two straight lines with the square on half their difference = the square on half their sum. Rect. AD, DB and sq. on CD = sq. on CB. EUCLID'S ENUNCIATION II. 5. 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. Cons. (Euclid's Method).-On CB describe a sq. draw its diagonal BE. Draw DG || to EC cutting BE. Through H the point of section draw KLM || to AB. Through A draw AK || to CL. Proof.-Comp. CH= Comp. HF (I. 43). AL CM DF, .. AH (AL+CH)+LG=CH+ DF + LG =sq. on CB. Cons. (Another Method).-Describe sq. on CB (half the line). From DG drawn to CE cut off DH= || Complete the rect. AD, DB. ᎠᏴ, Proof.-AH (AL + CH) + LG = CH + DF + LG, =sq. on CB. EUCLID'S ENUNCIATION II. 6, THEOREM 6. 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. Rect. AD, DB and sq. on CB = sq. on CD. Cons. (Euclid's Method).—Describe sq. on CD, Proceed as in II. 5. Proof.-As in II. 5. (The part of rect. AD, DB outside of the square on CD = one of the complements.) Another Method. In constructing rect. AD, DB cut off DM = DB from DF a side of the square. Proof. As in II. 5. AL CH = HF, AM (AL + CM) + LG = HF + CM + LG =sq. on CD. EUCLID'S ENUNCIATION II. 8, THEOREM 8. If a straight line be divided into any two parts, four times the rectangle contained by the whole line and one of the parts, together with the square on the other part, is equal to the square on the straight line, which is made up of the whole and the first part. 4 (rect. AB, BC) and sq. on AC = sq. on AD (AB + BC). Cons. (Euclid's Method).-Produce AB to D making BD = BC, On AD describe a square. Complete the figure as in preceding prop. drawing two Is each way. Proof.-CK BN GR=RNJ = A C B D } I. 34 and I. 36 G K M N X P R O CK=RN I. 43 .. The four small squares are equal. For the same reasons the four small rects. are equal. E 11 L F Rect. AB, BC = AK = 1 small rect. and 1 small square. .. 4 (rect. AB, BC) = 4 small rects. and 4 small squares. = gnomon AOH. Gnomon AOH and XH (i.e. sq. on AC) =sq. on AD. Cons. (Another Method.)-Produce AB to D, making way. In DF cut off DN, NO each = - BC. Complete figure as in previous prop., drawing two ||s each Proof. The four small rects. and the four small squares are equal (cons. and I. 34). Rect. AB, BC, i.e. AK = 1 small rect. and 1 small sq. .. 4 (rect. AB, BC) 4 small rects. and 4 small sqq. .. (4 rect. AB, BC) and square on AC, i.e. XH = 4 small sqq., and rects. together with XH = sq. on AD. LECTURE XX. PROPOSITIONS 7, 9, AND 10. PROPOSITION VII. is thus enunciated by Euclid: 'If a straight line (AB) be divided (at C) into any two parts, the squares on (AB and BC) the whole line, and one of the parts, are equal to twice the rectangle (AB, BC) contained by the whole and that part, together with the square on (AC) the other part.' Construct a figure similar to those in the preceding propositions. Then from the demonstration previously given we know that the rectangles AG, BF each rect, AB, BC, = You will notice that CG is common to both of the rectangles AG and BF. There- To each of these equals let us add KF (the square on AC), which completes the figure. Then we have this equation: Twice the rectangle AB, BC and the square on AC (KF): the square on AB (gnomon and KF), and square on BC.Q.E.D. = Now, instead of considering AC, CB as parts of a divided line let us consider that two straight lines are given, equal in length to AB and BC—then AC is clearly their difference, and the proposition may be enunciated thus : The square on the difference of two straight lines (sq. on AC), together with twice the rectangle contained by these lines (2 AB, BC) = the sum of the squares on the two lines (squares on AB, BC). From II. 4 (second enunciation) we learn that 'The square on the sum of two lines the sum of the squares on those two lines together with twice the rectangle contained by them." That is, Sq. on difference and twice the rect. contained by two lines = sum of sqq. on two lines (II. 7). Sq. on sum of two lines, less twice the rect. contained by two lines = sum of sqq. on two lines (II. 4). Adding these two equations (ax. 2), we infer that, Sq. on diff. and sq. on sum of two lines = twice the sum of the sqq. on two lines. This corollary will be found to include props. 9 and 10; enunciated by Euclid as follows (a. II. 9): If a straight line (AB) be divided into two equal and also into two unequal parts (at C and D), the squares on the two unequal parts (sqq. on AD, DB) are together double of the squares on half the line, and on the line between the point of section. Sqq. on AD, DB = twice the sqq. on AC and CD. Supposing AC and CD to be the two given lines, then since CB = AC (cons.), DB is their difference and AD their Hence by the corollary given above we have :The squares on AD the sum, and DB the difference twice the sum of the squares on the given lines AC, CD. sum. ·-- = (B. II. 10.) If a straight line AB be bisected at C and produced to any point D, the square on the whole line thus produced (AD), and the square on the part of it produced (BD), are together double of the squares on half the line bisected (AC), and of the line made up of the half and the part produced (CD).' Sqq. on AD, ᎠᏴ = twice the sqq. on AC, CD. Let AC and CD be the two given lines, then AD is their sum, and BD their difference (CB = AC cons.). Hence by this corollary |