Therefore, it is proved, as required, that

The

squares on AB and BC twice the rectangle AB, BC, together with the square on AC.

Wherefore,

If a straight line be divided, &c.

Exercise.

Q. E. D.

Prove the other case in the above Proposition, that

The squares on AB and AC = twice the rectangle AB, AC, together with the square on BC.

PROP. 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 line and that part.

C.

Let AB be a straight line divided into any two parts in

CONSTRUCTION.-1. Produce AB to D, making BD = BC (post. 2 and I. 3).

2. Upon AD describe the square AEFD (I. 46).

3. Complete the figure by the parallels cutting the diagonal, each way, in K and P (I. 31).

PROOF.-1. Because BD = BC (cons.), and = KN (I. 34), therefore BC and BD each KN (I. 34), therefore GK= KN (ax. 1), and similarly PR = RÒ.

2. Next, because CB = BD and GK = KN, therefore the rect. CK the rect. BN, and the rect. GR

KO (I. 36).

the rect.

3. But because the rect. CK = the rect. KO (I. 43), therefore the rect. BN = the rect. GR, and therefore the four rectangles CK, BN, GR, and KO= each other, therefore also they are, together, four times any one of them, as CK.

=

4. Next, because BD = BC (cons.), and because BD=BK (II. 4, Cor.) CG (I. 34), therefore BC=CG (ax. 1); and because BC=GK (1.34)=GP (II. 4, Cor.), therefore CG = GP (ax. 1); and therefore the rect. AG the rect. MP (I. 36);

and because PR = RO, therefore the rect. PL = the rect. RF (I. 36). But because the rect. MP the rect. PL (I. 43) therefore these four rectangles AG, MP, PL, and RF = each other, and therefore they are together four times any one of them, as AG.

5. And because the four rectangles CK, BN, GR, and KOJ

therefore the eight rectangles) making the gnomon AOH

=

(four times the rect. CK,

four times the rect. AK.

And because AK is the rect. AB, BC, since BK =

BD,

therefore four times the rect. AB, BC=four times the rect. AK,

and therefore four times the rect. AB, BC, with XH

= the gnomon AOH,

=

{

the gnomon AOH with XH,

and because XH = the square on XP=AC,

(the gnomon AOH with the square on AC,

= the whole figure AEFD, the square onAD(cons.), (the square on AB and BC together.

=

Therefore, it is proved, as required, that

Four times the rectangle AB, BC, together with the square on AC the square on AD, i.e. on AB and BC together.

Wherefore,

If a straight line be divided, &c.

Q. E. D.

N.B.-It will assist the learner to notice the following steps in this Proposition:

a. To prove that the four rectangles CK, BN, GR,
and KO = each other, and that together they are four
times any one of them, as CK (as in 1, 2, 3).
b. To prove that the four rectangles AG, MP, PL,
and RF each other, and that together they are
four times any one of them, as AG (as in 4).

c. To combine these conclusions is the proof of the
Proposition itself (as in 5).

F

PROP. 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 AB be a straight line divided into two equal parts in C, and into two unequal parts in D.

CONSTRUCTION.-1. From C draw CE at right angles to AB and AC or CB (I. 11, and 3).

2. Join EA and EB.

3. Through D draw DF parallel to CE (I. 31), meeting EB in F.

4. Through F draw FG parallel to AB (I. 31), meeting CE in G, and join AF.

PROOF.-1. Because AC CE (cons.), therefore the angle CEA = the angle CAE (I. 5); and because ACE is a right

angle (cons.), therefore the angles CEA and CAE are together a right angle (I. 32); and since they are equal to each other, therefore each angle CEA and CAE= half a right angle.

Similarly, each angle CEB and angle, and therefore the angles AEC i.e. the angle AEB = a right angle.

CBE = half a right and CEB together,

2. Next, because in the triangle EGF the angle GEF=the angle CEB (note 2 def. 15) = half a right angle, and because the angle EGF is a right angle, since the angle EGF = the angle GCB (I. 29)= the angle ECB (note def. 15) = a right angle (cons.), therefore the remaining angle EFG = half a right angle (I. 32), and therefore the side EG = the side GF (I. 6).

3. Again, because in the triangle FDB the angle FBD = the angle EBC, as above = half a right angle; and because the angle FDB is a right angle (cons.); therefore the remaining angle BFD= half a right angle the angle FBD, and therefore the side DF the side DB (I. 6).

4. Because in the triangle AEC the side AC = the side CE (cons.), therefore the square on AC the square on CE, and therefore the squares on AC and CE together double the square on AC.

=

But because the square on AE = the squares on AC and CE (I. 47), therefore the square on AE double the square on AC (ax. 1).

5. Because in the triangle GEF it has been proved that the side EG = the side GF, therefore the square on EG = the square on GF, and therefore the squares on EG and GF = double the square on GF.

But because the square on EF = the squares on EG and GF (I. 47), therefore the square on EF

on GF.

double the square

And because GF = CD (I. 34), therefore the square on EF= double the square on CD (ax. 1).