COR. If a Line be equally divided, the Rectangle under the Segments is a Square, and is equal to the Square of either Segment.

Hence, the Square of a whole Line is equal to four times the Square of half the Line. 4X4X4 8x8,64.

THEOREM V.

If a Right Line be bifected, and alfo cut unequally at pleafure; the Rectangle, under the two unequal parts added to the Square of the intermediate part, i. e. the difference between the equal and unequal, is equal to the Square of half the Line.

Let AB be cut equally in C, and unequally in D. Then, the Rectangle under AD and

DB, added to the Square of CD, is equal to the Square of AC or CB.

On CB, half AB, defcribe the CBEF. Draw DG parallel to CF; make DH qual DB, and draw KH parallel to

AB, and AK parallel to CF.

F

G

K

H

5

D

DEM. Now CFEB is a Square (Con.) and AC=CB- Hyp.

DH-Def. 7.
DB-Ax.3.

wh. BE AC (Ax. 3.) & AK, CI, are cach But DH DB (Con.) wh. AK & CI are each Th. the Rect. AKIC is equal to the Rect. DGEB. - Ax. Add, to both, the Rect. FD; &, AI+FD=FD+GB - 6. but AI+ID,eq. AH,—ADB□, & FGHI is the □ of CD; for, DG is parallel to CF, and are each equal to CB-Con. alfo, KH is parallel to AB, and CI is equal to DB; confequently, IF and GH are each equal to CD.- Ax. 7. Th. the Rect. AH+ the Square FH the Square CFEB.-2. į. ẹ. ADXDB+CD□=AC, or CB .

Let

Let the whole Line AB be 16; AC CB each =8; let CD be 3, and DB 5; then AC+CD, =AD,=11, Then, the Rect. AH, i. e. ADB, or 11x5=55 +the Square FH, i. e. CD, or 3×3=9 the Square CFEB, i. e.

CB □, or 8×8=64

If a Right Line be divided into two equal Parts, and then produced at pleasure, or another Line be added; the Rectangle contained under the whole compounded Line and the Part added, together with the Square of half the given Line, is equal to the Square of the half Line and the Part added, in one Line.

On CD, defcribe the Square CDEF; and on AD the Rectangle AKLD; by drawing AK parallel to CE, and KL to AD, making DL equal to BD. Draw BG parallel to DE, and join FD.

DEM. The Rect. AI=1B (Ax.) and IB=GL (19. 1.)
therefore, AI is equal to GL. -
Ax. 3.
Conf. the Rect. AI+IB+BL=IB+BL+GL - Ax. 6.
And, if to both, be added the Square IFGH, of CB.
The Rect. AKLD+IFGH=CFED.

i. e. ADB, or ADXDB+CB¤=CD¤.

Ax. 2.

O

Or it may be demonftrated as the foregoing. Produce BA, and make AE equal to BD.

Then is ED bifected, in C, and unequally cut, in B. Wherefore, the Rect. EBD, under the unequal parts, (equal ADB) i. e. EBxBD,+CB□=CDп.

Let AB be 12, bifected in C, AC, 6, and CB, 6, and let BD, the part added, be 4.

Then, the Rect. AKLD, i. e. ADX BD, or 16x 4 64 the Square IFGH, i. e. CBO,

the Square CFED, i. e. CD¤,

or 6×6= 36

or 10x10=100

THEOREM VII.

If a Line be divided, equally or unequally, at pleafure; the Square of the whole Line, added to the Square of either Segment, is equal to two Rectangles, under the whole Line and that Segment, together with the Square of the other Segment.

Let AB be divided, any how, in the point E. A 3 Then, the Square of A B, added to the Square of AE, is equal to two Rectangles under AB and that Segment, added to the Square of EB, the other Segment.

Defcribe the Square ABCD; through E, draw' EF parallel to AD; make AG equal AE, and draw GH, parallel to AB; and join AC.

Ax. 2.

DEM. The Rect. DI=IB (19.1.) add GE to both; &DE=GB But, the Rectangle DE+GB=DI+IB+2GE; - Th.3. and, if FH, i. e. the Square of EB, be added; they are equal to the Square ADCB+GE. - But, the Rectangles DE,GB, are under the whole Line, AB, and the Segment AE; for, AD=AB and AG=AE-Con. Therefore 2 BAXAE÷EB□=AB□+AE,

Allo, DH+FB+GE=DB4FH;

or 2ABE+AED=AB+EB. Y

Let

5

Let the whole Line, AB, be 8; let AE be 3, and EB 5.
Then the Square ABCD, of AB, i. e. 8x8 64

the Square AEIG, of AE, i. e. 3x3 9

=73

But, the two Rects. DE, GB, i. e. 2 AB×AE, 8×3=48

the Square IFCH, of EB,

5×5=25 is alfo =73

E

6

Α

If a Right Line be divided, any how, in two Parts; four Rectangles under the whole Line and either Segment, added to the Square of the other Segment, is equal to the Square of a Line, comRounded of the whole Line and the Segment first

C

3

B

3

H

taken.

M Let AB be divided, at pleafure, in C;

and, if CB be the Segment taken,

K make BD equal to BC.

Then, four Rectangles under AB and CB added to the Square of the Segment AC, D will be equal to the Square of AD.

On the whole Line, AD, construct the Square AEHD. Draw CF and BG, parallel to AE; make AI equal AC, and AL equal to AB, and draw IK, LM, parallel to AD.

DEM. Now CF & BG are parallel to AE; & IK,LM, to AD;
wherefore, FG, GH, HM, & MK, &c. are each equal CB;
confequently, FP, GM, OQ, and PK are equal Squares.
And, EO, LN, NB, and QD are equal Rectangles.
But the Rect. EO+FP=ABC, or ABXBC.
Conf. the four Rectangles EO, LN, &c. added to the four
Squares F P, GM, &c. are equal to four times ABXBC;
and, if the Square IC (of AC) be added, they are all
equal to the Square AEHD, of the Line, AD, com-
pounded of AB and the Segment CB, equal BD. - Ax. 2.
i. c. 4ABC or 4 times ABX BC+AC¤=AD¤,

Let AB be 9, divided at C, in 6 and 3.

and if BD be equal CB, then AD will be equal 12.

1

Then, the Rect. ABC, or ABXBC, i. e. 9x3=27.
Conf. 4 Rectangles, ABXBC is equal to 108

the Square of AC,

6x6

36

If a Right Line be divided, into two equal and two
unequal Parts; the two Squares of the unequal
Parts are, together, double the Square of half the
Line, together with the Square of the intermediate
Part.

Let AB be bifected in C, and cut unequally in D.

Then, the Square of AD, added to the Square of DB; is equal to twice the Square of AC, added to twice the Square of CD, the difference between AC and AD.

F

A

7

C2D 5

B

Draw CE perpendicular to AB, and equal to AC or CB.
Join AE and EB, and draw DG parellel to CE.
Through G, where DG cuts EB, draw FG parallel to AB;
and, laftly, draw AG,

3. 10. I.

DEM. Now fince ACE is a right Angle, & CE is equal AC,
the Angles CAE and AEC are half right. C.
And AE AC□+СE¤; i. e. equal 2 AC□.-20. 1.
for, ACE is a Right Angle, and AC=CE.

Con.

And, because CG is a Rectangle, FG=CD;
and FE is alfo equal CD; for EFG is a Right angle;
FEG (eq. FEA) is half right; th. the Angle FEG EGF
confequently, FE is equal to FG (equal CD) C. 3. 9. I.
and EGEFO+FGO,2 CD.

Y 2

20. I.

Now,