PROP. XXXII. 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.

H

For each exterior angle DEF, with its adjacent interior 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 Proposition immediately preceding, were

B

K

D

E

F

abated, to form the aggregate of the interior angles. Cor. If the figure has a re-entrant angle BCD, the an

Schol. The amount of the exterior angles might be deduced from the successive deflections which a side would make before it has returned to its first position. Thus, in the first case, AF makes a complete circuit, changing into the positions EG, DH, CI, BK, and finally into AF again. But in the second case, AG, after making similar deflections, turns backwards at C from the position DK to CL.

PROP. XXXIII. 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 and BC are equal and parallel to DC and DA.

For all the angles of the figure being equal to four right angles (I. 31. cor.), and the opposite angles being mutually equal, each pair of adjacent angles

B

C

must be equal to two right angles. Wherefore ABC and BCD are

equal to two right angles, and the A

D

lines AB and DC (I. 22. cor.) 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. But (I. 26.) the parallel sides of the figure are also equal.

Cor. Hence a quadrilateral figure contained by right angles has its opposite sides equal and parallel.

PROP. XXXIV. PROB.

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

From the point B, to draw a perpendicular to AB, without producing that line.

In AB take any point C, and on BC (I. 1. cor.) describe an isosceles triangle BDC, produce CD till DF be equal to it; and BF being joined, is the perpendicular required.

For, since by construction DF is equal to CD or BD, the triangle BDF is isosceles, and (I. 10.) the angle DBF equal to DFB; whence the angle CDB,

being equal (I. 30.) to the interior angles

DBF and DFB, is double of DBF, or

D

A

B

the angle DBF is half of CDB. But the triangle BDC being isosceles, the angle CBD is equal to BCD; consequently the angles DBF and DBC are the halves of the vertical and base angles of BDC, and therefore (I. 30.) the whole angle CBF is the half of two right angles, or it is equal to one right angle.

Schol. This problem, of which the construction may be slightly modified, is often more convenient in practice than the one given in the corollary to Prop. 5. of this Book.

PROP. XXXV. PROB.

On a given finite straight line, to construct a

square.

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

construct.

From the extremity B draw, by the last proposition, BC perpendicular to BA and equal

to it, and, from the points A and C with the distance BA or BC describe two circles intersecting each other in the point D, join AD and CD; the quadrilateral figure ABCD is the square required.

B

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. XXXVI. PROB.

To divide a given straight line into

of equal parts.

any number

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

E N

C

P

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. 23.), cutting AB in H, I, K, and L: AB is divided at these points into five equal parts.

M

AHIKL B

For (I. 23.) 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. 22.); 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. 20.) equal, and consequently AH is equal to DM. In the same manner, the triangle ADH is proved to be equal to EFN, to 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. 26.). Wherefore the several segments AH, HI, IK, KL, and LB, into which the straight line AB is divided, are all equal to each other.

Scholium. The construction of this problem may be facilitated in practice, by drawing from B in the opposite direction a straight line parallel to AC, and repeating on both of them portions equal to the assumed segment AD, but only four times, or one fewer than the number of divisions required; then joining D, the first section of AC, with the last of its parallel, E with the next, and so on till G, which connecting lines are (I. 27.) all parallel, and consequently the former demonstration still holds.