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If a straight line be bisected, and produced to any point, the square on the whole line thus produced, and the square on the part of it produced, are together double of the square on half the line bisected, and of the square on the line made up of the half and the part produced.

Let AB be bisected in C and produced to D.

Then shall sqs. on AD, DB be double sqs. on AC, CD.

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Constr. From C draw CE | AB, (i. 11), make CE = AC

or CB, (i. 3) join AE, EB; through E draw E F || AB,

and through D draw DF || CE. Then : EF meets || s CE, FD, .. int. Ls CEF, EFD = two rt. Ls;

(i. 29) i. _s BEF, EFD are less than two rt. Zs, and

:: EB, FD will meet, if produced ; (ax. 12)

let them meet in G, and join AG. Dem. .. AC = CE, :: CAE = Z CEA; (i. 5)

and LACE is a rt. L; .. each of Ls CAE, CEA is half a rt. L.

(i. 32) Similarly, each of <s CEB, CBE is half a rt. L.

.. LAEB is a rt. L. And :: EBC is half a rt. L,.. DBG is also half a rt. L;

(i. 15) but _BDG is a rt. L, being = alt. _ DCE; (i. 29) .. rem. _ DGB is half a rt. L, and is .. = _DBG ; .. BD = DG.

(i. 6) Again, : EGF is half a rt. L, and _EFD is a rt. L, being = the opposite _ ECD,

(i. 34) :. rem. _FEG is half a rt. L, and is .. = LEGF; .. GF= FE.

(i. 6) And •: EC = CA, .. sq. on EC = sq. on CA; .. sqs.

on EC, CA are double of sq. on CA; but sq. on EA = sqs. on EC, CA; .. sq. on EA is double

of sq. on CA. And : GF EF, .. sq. on GF = sq. on EF, .. sqs. on

GF, FE are double of sq. on EF ; but sq. on EG = sqs. on GF, FE; .. sq. on EG is double

of sq. on FE; and EF CD, (i. 34) .. sq. on EG is double of sq. on

=

CD;

and it was shown that sq. on EA is double of sq. on CA; .. sqs. on EA, EG are double of sqs. on AC, CD;

but sq. on AG = sqs. on EA, EG; (i. 47)

i. sq. on AG is double of sqs. on AC, CD; but sqs. on AD, DG = sq. on AG; .. sqs. on AD, DG are double of sqs. on AC, CD;

but DG = DB ;
i sqs. on AD, DB are double of sqs. on AC, CD.

Therefore, if a straight line, &c. Q. E. D. [The demonstration corresponds closely with that of the preceding proposition, and the successive steps are these :(1) ZAEB is a rt. 2; (2) BD = DG; (3) GF = FE ; (4) sq. on EA is double sq. on CA; (5) sq. on EG is double sq. on CD; (6) sq. on AG is double sqs. on AC, CD; (7) sqs. on AD, DB are double sqs. on AC, CD.

The corresponding algebraical result is as follows :-Let AB contain 2a, and BD x units. Now (2a + x)2 = 4a2 + 4ax + x2; add za to each of these = s, and (2a + x)2 + x2 = 4a? + 4ax + 2x2 = 2a2 + 2(a2 + 2ax + x*)= 2a2 +2(a + x)2; that is, if a number be divided, &c.

The proper construction for the solution of a geometrical problem may often be indicated by employing algebraical symbols. The student who has been able to solve a fair number of the riders up to this point may with advantage consider the following problem, illustrating this remark: • To divide a given line into two parts so that the square on one part may be double the square on the other.' Let the length of the line AB be a, suppose it be divided as required in D, and let DB = x. Then since sq. on AD= twice sq. on BD .. (a – x)2 = 2x2. A solution of this quadratic is x= al 2-1). Now a v2 indicates the diagonal of the sq. on AB (for twice the sq. on a side = sq. on the diagonal, by i. 47), hence the following construction is suggested by the solution. Produce AB to C making BC = AB. On BC describe a square, and with centre C, radius = diagonal of sq., describe a circle cutting AB in D: D is the point required.

For by prop. 9, sq. on AD with sq. on DC are double sqs. on DB and BC. But sq. on DC is double sq. on BC, by constr., .. sq. on AD is double sq. on DB. Again, take this problem : "To produce a given straight line so that the square on the whole line thus produced may be double the square on the part produced.' As before, let AB = a, BD = x, then since sq. on AD= twice sq. on BD, (a + x)2 = 2x2. This quadratic solved gives x = a(N2+1) as one solution, and this value suggests the construction, namely : produce AB to C making BC = AB; on BC describe a square : with centre C and radius = diagonal of sq. describe a circle, cutting AC produced in D: D is the pt. required. For by prop. 10, sqs. on AD, DC are double sqs. on DB, BC. But sq. on DC (= sq. on diagonal) is double sq. on BC, .. sq. on AD is double sq. on BD.

The thoughtful reader will have observed a striking correspondence between the solutions of these two problems and the indicating equations, and he will also have noticed that each of the quadratics has two solutions. We cannot dwell further on this remarkable point at present : the student may return to it with great profit at a more advanced stage of his reading. It is sufficient to note the use of algebraical symbols in suggesting the solution of problems of this nature. Observe, in suggesting only :-see the note at the beginning of Book ii.]

PROPOSITION XI. PROBLEM.

To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts may be equal to the square on the other part.

Let AB be the given st. line. It is required to divide AB into two parts, so that rect. contained by AB and one part may

= sq. on the other part.

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Constr. On AB describe sq. AD; (i. 46) bisect AC at

E, (i. 10) and join BE ; produce CA to F, make EF = EB, and on AF describe

sq. FH.

Then shall AB be divided at H so that rect. AB, BH

= sq. on AH. Produce GH to meet CD in K. Dem. :: AC is bisected in E and produced to F, ..

rect. CF, FA with sq. on AE sq. on EF; (ii. 6) but EF = EB, :, rect. CF, FA with sq. on AE = sq. on

EB;

but sqs. on BA, AE sq. on EB, (i. 47); .. rect. CF, FA with sq. on AE = sqs. on BA, AE; take away sq. on AE from each of these =s;

.. rect. CF, FA = sq. on BA.

But FK is rect. CF, FA, for FA = FG ; and AD is sq. on AB ;

:: FK = AD; take

away the common part AK,

.. rem. FH = rem. HD;
but HD = rect. AB, BH, for AB = BD;

and FH is sq. on AH;

., rect. AB, BH sq. on AH, Therefore, the given straight line is divided as required.

Q. E. F. 1. Show that the squares on the whole line and one part are equal to three times the square on the other part.

2. Show that the rectangle contained by the sum and difference of the parts is equal to the rectangle contained by the parts.

3. If CH produced meets BF at L, the angle CLB is a right angle.

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