Book I.

a 42. 1.

b 44. 1.

Join DB, and describe the parallelogram FH equal to the triangle ADB, and having the angle HKF equal to the angle E; and to the straight line GHb apply the parallelogram GM equal to the triangle DBC, having the angle GHM equal to the angle E. Because the angle E is equal to each of the angles FKH, GHM, the angle FKH is equal to GHM; add to each of these the angle KHG; therefore the angles

c 29. 1.

d 14. 1.

e 30. 1.

FKH, KHG are equal to the angles KHG, GHM; but FKH, KHG are equal to two right angles; therefore also KHG, GHM are equal to two right angles; therefore KH is in the same straight lined with HM. Because the straight line HG meets the parallels KM, FG, the alternate angles MHG, HGF are equal: add to each of these the angle HGL; therefore the angles MHG, HGL are equal to the angles HGF, HGL; but the angles MHG, HGL are equal to two right angles; wherefore also the angles HGF, HGL are equal to two right angles; therefore FG is in the same straight lined with GL. Because KF is parallel to HG, and HG to ML, KF is parallel to ML; but KM, FL are parallels; wherefore KFLM is a parallelogram. Because the triangle ABD is equal to the parallelogram HF, and the triangle DBC to the parallelogram GM, the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM; therefore the parallelogram KFLM has been described equal to the given rectilineal figure ABCD, having the angle FKM equal to the given angle E. Which was to be done.

COR. From this it is manifest how to a given straight line to apply a parallelogram, which shall have an angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure, viz. by applying to the given straight line a parallelogram equal to the first triangle ABD, and having an angle equal to the given angle.

Book I.

PROP. XLVI. PROB.

TO describe a square upon a given straight line.

Let AB be the given straight line; it is required to describe a square upon AB.

EB

C

E

d 34. 1.

From the point A drawa AC at right angles to AB, and a 11. 1. makeb AD equal to AB, and through the point D draw b 3.1. DE parallel to AB, and through B draw BE parallel to c 31. 1. AD; then ADEB is a parallelogram; therefore AB is equald to DE, and AD to BE. But BA is equal to AD; therefore the four straight lines BA, AD, DE, are equal to one another, and the pa- D rallelogram ADEB is equilateral. It is likewise rectangular; for the straight line AD meeting the parallels AB, DE, makes the angles BAD, ADE equale to two right angles; but BAD is a right angle; therefore also ADE is a right angle. A Now the opposite angles of parallelo

e 29. 1.

B

grams are equald; therefore each of the opposite angles ABE, BED is a right angle; wherefore the figure ADEB is rectangular; and it has been demonstrated that it is equilateral; it is therefore a square, and it is described upon the given straight line AB. Which was to be done.

COR. Hence every parallelogram that has one right angle has all its angles right angles.

PROP. XLVII. THEOR.

IN any right angled triangle the square which is described upon the side subtending the right angle is equal to the squares described upon the sides which contain the right angle.

Book I.

a 46. 1.

b 31. 1.

Let ABC be a right angled triangle having the right angle BAC; the square described upon the side BC is equal to the squares described upon BA, AC.

On BC describe the square BDEC, and on BA, AC the squares GB, HC; and through A drawb AL parallel to BD or CE, and join AD, FC; then, because each of the angles BAC, BAG is a right

c 25. Def. angles, the two straight
lines AC, AG upon the
opposite sides of AB,
make with it at the point F
A the adjacent angles e-
qual to two right angles;
therefore CA is in the
same straight lined with
AG; for the same reason,
AB and AH are in the
same straight line. Now
because the angle DBC is

d 14. 1.

e 10. Ax. equale to the angle FBA, each of them being a right

2. Ax.

g 4. 1.

h 41. 1.

k 6. Ax.

angle, add to each the an

B

D

H

L E

K

gle ABC, and the whole angle DBA will be equal to the whole FBC; and because the two sides AB, BD are equal to the two FB, BC, each to each, and the angle DBA equal to the angle FBC, therefore the base AD is equals to the base FC, and the triangle ABD to the triangle FBC. But the parallelogram BL is doubleh of the triangle ABD, and the square GB is double of the triangle BFC; therefore the parallelogram BL is equal to the square GB. In the same manner, by joining AE, BK, it may be demonstrated that the parallelogram CL is equal to the square HC. Therefore the whole square BDEC is equal to the two squares GB, HC : and the square BDEC is described upon the straight line BC, and the squares GB, HC upon BA, AC. Wherefore the square upon the side BC is equal to the squares upon the sides BA, AC. Therefore, in any right angled triangle, &c. Q. E. D.

OF GEOMETRY.

PROP. XLVIII. THEOR.:

IF the square described upon one of the sides of a triangle be equal to the squares described upon the other two sides of it, the angle contained by these two sides is a right angle.

If the square described upon BC, one of the sides of the triangle ABC, be equal to the squares upon the other sides BA, AC, the angle BAC is a right angle.

D

Book I.

From the point A drawa AD at right angles to AC, and a 11. 1. makeb AD equal to BA, and join DC. Then because DA is b 3.1. equal to AB, the square of DA is equal to the square of AB. To each of these add the square of AC; therefore the squares of DA, AC are equal to the squares of BA, AC. But the square of DC is equal to the squares of DA, AC, because DAC is a right angle; and the square of BC, by hypothesis, is equal to the squares of BA, AC; there- B fore the square of DC is equal to the

A

c 47. 1.

C

square of BC; therefore the side DC is equal to the side BC.
Because the side DA is equal to AB, and AC common to
the two triangles DAC, BAC, and the base DC equal to the
base BC, the angle DAC is equal to the angle BAC. But d 8. 1.
DAC is a right angle; therefore also BAC is a right angle.
Therefore, if the square, &c. Q. E. D.

The following proposition is placed here, because it is more connected with the first book than with any other. It is useful for explaining the nature of Hadley's sextant, and, though involved in the explanations usually given of that instrument, it has not, I believe, been hitherto considered as a distinct geometric proposition, though very well entitled to be so, on account of its simplicity and elegance, as well as its utility.

Book 1.

a 16. 1.

b 4. Ax.

c 13. 1.

€ 32. 1.

PROP. A. THEOR.

IF an exterior angle of a triangle be bisected, and also one of the interior and opposite angles, the angle contained by the bisecting lines is equal to half the other interior and opposite angle of the triangle.

Let the exterior angle ACD of the triangle ABC be bisected by the straight line CE, and the interior and opposite ABC by the straight line BE; the angle BEC is equal to half the angle BAC.

The lines CE, BE will meet; for since the angle ACD is greatera than ABC, the half of ACD is greater than the half of ABC, that is, ECD is greater than EBC. Add

E

A

B

C

ECB to both, and the two angles ECD, ECB are greaterb than EBC, ECB. But ECD, ECB are equal to two right angles; therefore ECB, EBC are less than two right angles, and therefore the lines CE, BE must meet on the same side d Cor. 29.1. of BC on which the triangle ABC isd. Let them meet in E. Because DCE is the exterior angle of the triangle BCE, it is equale to the two angles CBE, BEC, and therefore twice the angle DCE, that is, the angle DCA is equal to twice the angles CBE, BEC. But twice the angle CBE is equal to the angle ABC; therefore the angle DCA is equal to the angle ABC, together with twice the angle BEC; and the same angle DCA, being the exterior angle of the triangle ABC is equal to the two angles ABC, CAB; wherefore the two angles ABC, CAB are equal to ABC and twice BEC. Therefore if ABC be taken from both, there remains the angle CAB equal to twice the angle BEC, or BEC equal to the half of BAC. Therefore, &c. Q. E. D.