PROP. XXXV. THEOR.

The angles round any rectilineal figure are together equal to twice as many right angles, abating four from the result, as the figure has sides.

For assume any point O within the figure, and draw straight lines OA, OB, OC, OD, and OE, to the several angular points. It is obvious, that the figure is thus resolved into as many triangles as it has B

A

D

sides, and whose collected angles must be, therefore, equal to twice as many right angles. But the angles at the bases of these triangles constitute the internal angles of the figure. Consequently, from the whole amount there is to be deducted the vertical angles about the point O, and which are (Def. 4.) equal to four right angles.

Cor. Hence all the angles of a quadrilateral figure are equal to four right angles, those of a pentelateral figure equal to six right angles, and so forth; increasing the amount by two right angles for each additional side.

PROP. XXXVI. THEOR.

The exterior angles of a rectilineal figure are together equal to four right angles.

The exterior angles DEF, CDG, BCH, ABI, and EAK of the rectilineal figure ABCDE are taken together equal to four right angles.

For each exterior angle DEF, with its adjacent inte

rior one AED, is equal to two

right angles. All the exterior angles therefore, added to the interior angles, are equal to

twice as many right angles as the figure has sides. Consequently the exterior angles are equal to the four right angles which, by the last Proposition, were abated, to form the aggregate of the interior angles.

I

B

D

E

F

Cor. If the figure has a re-entrant angle BCD, the angle

BCK which occurs in place of an exterior angle, must be taken away in forming the amount; for the corresponding interior angle BCD, in this case, exceeds two right angles by BCK. Hence the angles EFG, DEH, CDI, ABL, FAM diminished by BCK are equal to four right angles.

B

K

I

C

E

F

G

H

PROP. XXXVII. THEOR.

If the opposite angles of a quadrilateral figure be equal, its opposite sides will be likewise equal and parallel.

In the quadrilateral figure ABCD, let the angle at B be equal to the opposite one at D, and the angle at A equal to that at C; the sides AB, BC are equal and parallel to DC and DA.

For all the angles of the figure being equal to four right angles, (I. 34. cor.) and the opposite angles being mutually equal,

each pair of adjacent angles must be equal to two right angles. Wherefore ABC and BCD are

.A.

B

D

equal to two right angles, and the lines AB and CD (Cor, I. 25.) parallel; for the same reason, ABC and BAD being together equal to two right angles, the sides BC and AD, which limit them, are parallel.

Cor. Hence a rectangle, or right-angled quadrilateral figure, has its opposite sides equal and parallel.

PROP. XXXVIII. PROB.

To draw a perpendicular from the extremity of a given straight line.

From the point B, to draw a perpendicular to AB, with out producing that line.

In AB take any point C, and on BC (I. 1. cor.) describe an equilateral triangle CDB, on its side DB, another DEB; and on DE the side of this, a third equilateral triangle DFE; join the last vertex F with the point B; and BF is the perpendicular required.

A

C

D

Because the triangles CDB and DBE are equilateral, the angles CBD and DBE are each of them equal to two third parts of a right angle (I. 34. cor.); and the triangles BDF, BEF, having the sides BD, DF equal to BE, EF, and the side BF common, are (I. 2.) equal, and con

sequently the angles FBD and FBE are equal, and each of them the half of DBE. The angle FBD, being therefore one third part of a right angle, and the angle DBA two third parts, the whole angle FBC must be an entire right angle, or the straight line BF is perpendicular to AB.

PROP. XXXIX. PROB.

On a given finite straight line to construct a square.

Let AB be the side of the square which it is required to

For, by this construction, the figure has all its sides equal, and one of its angles ABC a right angle; which comprehends the whole of the definition of a square.

PROP. XL. PROB.

To divide a given straight line into any number of equal parts.

Let it be required to divide the straight line AB into a given number of equal parts, suppose five.

From the point A and at any oblique angle with AB draw a straight line AC in which take the portion AD, and repeat it five times from A to C, join

CB, and from the several points of section D, E, F, and G draw the parallels DH, EI, FK, and GL, (I. 26.) cutting AB in H, I, K, and L: AB is divided in these points into five equal parts.

E

N

M

AHIK LB

For draw DM, EN, FO, and GP parallel to AB. And because DH is parallel to EM, the exterior angle ADH is equal to DEM (I. 25.); and for the same reason, since AH is parallel to DM, the angle DAH is equal to EDM. Wherefore the triangles ADH and DEM, having two angles respectively equal, and the interjacent sides AD, DE, are (I. 23.) equal, and consequently AH is equal to DM. In the same manner, the triangle ADH is proved to be equal to EFN, FGO, and GCP, and therefore their bases EN, FO, and GP are all equal to AH. But these. lines are equal to HI, IK, KL, and LB, for the opposite sides of parallelograms are equal (I. 29.). Wherefore the several segments AH, HI, IK, KL, and LB, into which the straight line AB is divided, are all equal to each other.