a

=

GF+FE2FE2CD2, fince EF= CD. Whence AE2 + EG2 = 2AC2+ 2CD2; But AE+EGa (AG2) = (aAD2+DG2)

b

2

- AD + DB2, fince DG = DB. Therefore AD2 + DB2 = 2AC2 + 2CD2. Q. E. D.

a

Upon AB defcribe the Square AC, and bifect the Side AD in E,and draw EB, and cut off from AE continued the part EF EB, and defcribe a Square upon AF; then fhall AH (=AG3) ABX BG. For continue out HG to I, then

a

the Rectang. DHEA

2

2

ΕΒ

е

BA+ EA. Therefore DH = BA Square AC. Take away the common Rectang. AI, and there will remain the Square AH GC; that is, AG2 = AB × BG. Q. E. D.

2

f

=

This Propofition cannot be explain'd in Numbers; for no Number can be fo divided, that the Product of the Whole into one of the Parts hall be equal to the Square of the other Part.

.PRO P.

PROP.

B

XII.

A If in an obtufe-angled
Triangle ABC, the Square
of the Side AC fubtending
the obtufe Angle ABC is
greater than the Squares ·
of AB, BC, the Sides con-
D taining the obtufe Angle, by
twice the Rectangle under

one of the Sides BC, containing the obtufe Angle ABC,
viz. that continued out, on which the Perpendicular
AD falls, and the Line BD taken without between
the Perpendicular AD and the obtufe Angle.

That is, ACCB2+2CBD + AB2.

b

For AC(CD+ADCB+ 2CBD 47. I. + BD2 + AD2 =) a CB + 2CBD + AB. 4.2.

Q. E. D.

SCHOLIU M.

2

Hence, If the Sides of an obtuse-angled Triangle ABC be known, we can easily find the Segment BD intercepted between the Perpendicular AD, and the Obtufe Angle ABC, as alfo the Perpendicular AD. As, Let AC be 10, AB 7, CB 5. Then AC 100, AB2 CB2 49, 25. Confequently AB+ CB74. Take this from 100, and there remains 262CBD. Whence CBD fhall be 13; which divided by CB 5, and there will arife 2 for BD: whence AD is found by Prop.47. 1.

--2

PROP. XIII.

In an acute-angled Triangle ABC, the Square of the Side AB fubtending the acute Angle ACB, is lefs than the Squares of the Sides AC, CB, comprehending the acute Angle ACB, by twice a Rectangle compre

hended

A

247. I. 7.2.

hended under one of the Side BC, about the acute Angle ACB, viz. on which the Perpendicular AD falls, and the Line DC taken within the Triangle from the Perpendicular AD to the acute Angle

D B ACB.

2

That is, AC+ BC2= AB2+ 2BCD.

2

ForAC+BC (AD2+ DC+BC="AD' +BD2+2BCD=) AB + 2BCD. Q. E.D.

a

SCHOLIU M.

Hence, likewife if the three Sides of the Triangle ABC be known, the Segment DC intercepted between the Perpendicular AD and the acute Angle ABC, as alfo the faid Perpendicular, may eafily be known.

Let AB be 13, AC 15, BC 14; take away AB2 (169) from AC+ BC, that is, from 225 + 196=421; and there remains 252 for 2BCD. Whence BCD fhall be 126. Divide this by BC 14, and there comes out 9 for DC; whence AD =√225 — 81 = 12.

PROP. XIV.

To make a Square ML equal to a given rights lined Figure A.

a

Make the Rectang. BD

b

a

A, and continue 2 45. I. out the greater Side DC to F, fo that CF CB. Bifect DF in G, from which, as a Center, de- b 10. I. fcribe the Circle FHD with the Distance GF, and continue out CB to meet the Circumference in H. Then fhall CH ML = A.

с

For draw GH; then is A"=DB =* DCF =
HC ML. Q. E.D.

GF GC

2

C.45. I.

conft.

End of the Second Book.