PROPOSITION VIII. THEOREM.

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 that part.

Let AB be divided into any two parts at C.

Then four times rect. AB, BC with sq. on AC

on the line made up on AB and BC together.

= sq.

Constr. Produce AB to D making BD = CB, (i. 3) on AD describe sq. AF, and complete the figures as in the preceding propositions.

Dem.

CB = BD, and CB = GK, BD = KN,

.. GK = KN; similarly PR = RO ;

hence, CB BD, and GK = KN,

.. rect. CK = rect. BN, and GR = RN; (i. 36)

but compl. CK = compl. RN, (i. 43) .. BN = GR ; .. the four rects. BN, CK, GR, RN are one another,

and are together quadruple one of them, CK.

Again, ... CB = BD, BD

=

and CB GK, that is

BK, that is = CG ;

= GP; .. CG =GP.

CG = GP, and PR = RO, .. rect. AG =

And
and PL= RF;

MP,

but compl. MP = compl. PL, (i. 43) .. AG = RF ; .. also the four rects. AG, MP, PL, RF are = one another, and are together quadruple of one of them, AG.

And it was shown that CK, BN, GR, RN are quadruple of CK; .. the eight rects. which make up the gnomon AOH are quadruple of AK.

But AK is rect. AB, BC,

BK= BC; .. four times

rect. AB, BC is quadruple of AK, and is therefore equal to the gnomon AOH.

To each add XH, which = sq. on AC;

(ax. 1)

.. four times rect. AB, BC, with sq. on AC, = gnomon AOH and sq. XH, which together make up the figure AF, the sq. on AD;

.. four times rect. AB, BC together with sq. on AC = sq. on AD, that is on the line made up of AB and BC together.

Therefore, if a straight line, &c.

Q. E. D.

[Corresponding result in algebra:-Let AB which contains a units be divided at C, and let AC contain x units; then since (a − x)2 = a2 — 2ax+x2, add 4ax to both sides and 4ax + (a− x)2 = a2 + 2ax + x2 = (a + x)3. Now BC contains a-x units, hence the analogous result is established.]

PROPOSITION IX. THEOREM.

If a straight line be divided into two equal, and also into two unequal parts, the squares on the two unequal parts are together double of the square on half the line, and of the square on the line between the points of section.

Let the st. line AB be divided into two equal parts at C, and into two unequal parts at D.

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

B

Constr. From C draw CEL AB, (i. 11) make CE=AC or CB, (i. 3) join EA, EB; through D draw DF || CE, meeting EB in F, through F draw FG || AB, and join AF.

Dem. ACCE, .. EAC AEC; (i. 5) and ZACE is a rt. 2. 4s CEA, CAE are together = a rt. L; (i. 32)

but they are one another, .. each of them is half a

rt. L.

Similarly each of the s CEB, EBC is half a rt. 4;

Again,
is int.

.. ▲ AEB is a rt. ▲.

GEF is half a rt., and EGF a rt. ▲, for it

ECB, (i. 29) .. rem. EFG is half a rt. ▲,

.. GEF = ▲ EFG, .. EG = GF.

(i. 6)

Again,.. for it is

=

ABE is half a rt. ▲, and FDB is a rt. ▲,

int. ECB, (i. 29) .. rem. ▲ BFD is half a rt. 2, .. DBF = 4. BFD, .. DF = DB. (i. 6) And

AC = CE, .. sq. on AC = sq. on CE, .. sqs. on AC, CE are together double of sq. on AC;

but sq. on AE is = sqs. on AC, CE ;

Again,

.. sq. on AE is double of sq. on AC.

(i. 47)

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

.. sq. on EF is double of sq. on GF;

(i. 47)

and GF = CD, (i. 34) .. sq. on EF is double of sq. on CD.

Now sq. on AE is double of sq. on AC, .. sqs. on AE, EF are double of sqs. on AC, CD ;

but sq. on AF = sqs. on AE, EF ;

.. sq. on AF is double of sqs. on AC, CD;

(i. 47)

(i. 47)

.. sqs. on AD, DF are double of sqs. on AC, CD;

and DF = DB;

.. sqs. on AD, DB are double of sqs. on AC, CD.

Therefore, if a straight line, &c.

Q. E. D.

[The demonstration is long, though its different steps are simple; it is shown (1) AEB is a rt. 4, (2) EG=GF, (3) DF = DB, (4) sq. on AE is double sq. on AC, (5) sq. on EF is double sq. on CD, (6) sq. on AF is double of sqs. on AC, CD, (7) sqs. on AD, DB are double sqs. on AC, CD.

The corresponding formula in algebra may thus be obtained. Let AB contain 2a units and let CD contain x units, so that AD contains a +x, and DB, a-x units. Now (a + x)2 = a2 + zax + x2; (a−x)2 = a2 −ax+x2; ... adding, (a + x)2 + (a − x)2 = 2a2 + 2x2: that is, if a number be divided, &c.]

1. The square on the sum of two straight lines to

gether with the square on their difference is double the squares on the two lines.

2. Divide a straight line into two parts so that the sum of their squares may be the least possible.

3. ABCD is a square and AC its diagonal. In AC a point E is taken. Show that the triangle, whose sides are equal to AE, EC, and the diagonal of a square described on BE, will contain a right angle.