Corollary 2.-All the exterior angles of any convex rectilineal figure are together equal to four right angles.

Let ABCDE be any rectilineal figure and let each of its sides be produced in order;

all the exterior angles so formed shall be together equal to four right angles.

Proof. Any interior angle ABC with its adjacent exterior angle ABK is equal to two right angles,

Prop. 13 therefore all the interior angles, together with all the exterior angles, are equal to twice as many right angles as

ABCDE has sides.

But by Corollary 1 all the interior angles, together with four right angles, are equal to twice as many right angles as the figure has sides;

therefore all the interior angles, together with all the exterior angles, are equal to all the interior angles, together with four right angles. Take away the common interior angles;

therefore all the exterior angles are together equal to four

1. What had been proved about the exterior angle of a triangle before Prop. 32?

2. What is the size of each of the exterior angles of an equilateral triangle? 3. If the five exterior angles of the figure ABCDE are all equal to one another, what is the size of each exterior angle?

4. AB is a given finite straight line; on AB describe an isosceles triangle ABC, making BC equal to AC; produce AC to D, making CD equal to AC or CB; join BD. Prove that the angle ABD is a right angle.

NOTES ON PROPOSITIONS 27-32.

Proposition 27.-This proposition is proved by Prop. 16 and Reductio ad absurdum. Euclid assumes that the lines AB and CD are in the same plane. Proposition 28.-This really contains two propositions, and the student should carefully note that the two things taken for granted in the enunciation are joined by the conjunction or. Either of the things taken for granted is quite sufficient to prove that the two lines are parallel. Both proofs depend on Prop. 27, the first using also Prop. 15, and the second using Prop. 13. Proposition 29.-This proposition is the converse of Props. 27 and 28. word "parallel" at the beginning of the enunciation is important to remember. The proposition is divided into three parts. The first part is proved by Reductio ad absurdum, and is noticeable for Axiom 12 being used in it for the first time in Euclid. The second part of the proof begins by assuming the first part; and the third part begins by assuming the second part.

The

Axiom 12 has been objected to because (1) it is not easy, and (2) it is not self-evident, whereas an axiom should be both easy and self-evident; and because (3) Axiom 12 being the converse of Prop. 17, if the converse to Axiom 12 needs proof, Axiom 12 also ought to be proved.

Many attempts have been made to improve on this part of Euclid, but none are satisfactory. The simplest axiom that has been suggested is the following, which is known as Playfair's Axiom :

Through a given point only one straight line can be drawn parallel to a given straight line.

This is the same as:- -Two straight lines which intersect cannot be both parallel to the same straight line. In this form Playfair's Axiom is the converse of Euclid's Prop. 30, and has therefore been objected to.

Assuming Playfair's Axiom instead of Axiom 12, the proof of Prop. 29 is as follows:-If the angle AGH is not equal to the angle GHD, through G draw PGQ making the angle PGH equal to the angle GHD; then by Prop. 27 PQ is parallel to CD; therefore AB and PQ which intersect at G are both parallel to CD; which is impossible by Playfair's Axiom. Therefore the angle AGH must be equal to the angle GHD.

Propositions 30 and 31.-Both these Props. are easily proved by Prop. 27. Proposition 32.-This is a most important theorem. It contains two distinct propositions, which also include the results of Props. 16 and 17. The first proof is a rider on Prop. 29; and the second proof, which commences by assuming the first proof, is a rider on Prop. 13.

They were not in Euclid

The two corollaries are important and difficult. originally, but were added by Simson. They may well be omitted till the student has finished the first book of Euclid. Of the two corollaries the first is the more important, because it enables us to calculate the size and magnitude of the interior angles of any regular rectilineal figure, e.g. :—

All the interior angles of a regular pentagon are equal to (2 × 5 4) = 6 right angles; therefore each angle of a regular pentagon is equal to & right angles.

MISCELLANEOUS RIDERS ON PROPOSITIONS 27-32.

1. Straight lines which are perpendicular to the same straight line are parallel to one another.

2. If a straight line meets two or more parallel straight lines and is perpendicular to one of them, it is perpendicular to all the rest.

3. If the diagonals of a quadrilateral figure bisect each other, the opposite sides of the figure are parallel.

4. If the arms of one angle are parallel to the arms of another angle, then the angles are either equal or supplementary.

5. If the arms of one angle are perpendicular to the arms of another angle, then the angles are either equal or supplementary.

6. ABC is an isosceles triangle, having the side AB equal to the side AC; the side BA is produced to D, and the exterior angle CAD is bisected by the straight line AX. Prove that AX is parallel to BC.

7. State and prove the converse of the last rider (No. 6).

8. Two straight lines which intersect cannot both be parallel to the same straight line.

9. Through a given point P draw a straight line that shall make with a given straight line AB an angle equal to a given angle C.

10. Prove that the three angles of a triangle are together equal to two right angles by joining the vertex to any point in the base.

11. Show how to divide a right-angled triangle into two isosceles triangles, and hence prove that the straight line drawn from the middle point of the hypotenuse to the right angle is equal to half the hypotenuse.

12. Draw a straight line at right angles to a given finite straight line from the extremity of it without producing the given line.

13. BAC is a given angle, and P is any point between AB and AC. Show how to draw through Pa straight line XPY, meeting AB in X and AC in Y and having PX equal to PY.

What difference will it make in the problem if P is (1) outside either AB or AC, or (2) on either AB or AC?

14. What is the size of each of the angles of (1) a regular hexagon, (2) a regular octagon, and (3) a regular dodecagon?

15. What is the size of each of the angles of a regular polygon of n sides? 16. Trisect a right angle.

17. ABC is any triangle, and the point 0 is taken within it, such that OA, OB, and OC are all equal. Prove that the angle BOC is double the angle

BAC.

18. ABCDE is a regular pentagon.

Prove that the lines AC, AD trisect the

angle BAE; and that ACD is an isosceles triangle having each of the angles ACD and ADC double the angle CAD.

PROP. 33.-Theorem.-The straight lines which join the extremities of two equal and parallel straight lines, towards the same parts, are themselves equal and parallel.

Let AB and CD be two equal and parallel straight lines, and let them be joined towards the same parts by the straight lines AC and BD;

AC shall be equal and parallel to BD

Proof.-(1) Because AB is parallel to CD, and BC falls on them, Hyp. therefore the angle ABC is equal to the alternate angle BCD. Pp. 29 (2) Again, because in the two triangles ABC, BCD

AB is equal to CD,

and BC is common to both triangles,

Hyp.

Because and the contained angle ABC is equal to the contained

therefore by Proposition 4 the triangle ABC is equal to the triangle BCD in all respects,

so that the base AC is equal to the base BD, and the angle ACB is equal to the angle CBD; but these are alternate angles;

therefore AC is parallel to BD.

Prop. 27

Therefore AC is both equal and parallel to BD. Q. E. D.

EXERCISE.

In the figure of Prop. 33 join AD and let AD and BC intersect in 0; prove

that AO is equal to OD and CO to OB.

PROP. 34.-Theorem.-The opposite sides and angles of a parallelogram are equal, and each diagonal of a parallelogram bisects its area.

Let ABCD be a parallelogram, and AC a diagonal of it;

the opposite sides and angles of ABCD shall be equal, and the diagonal AC shall bisect its area.

Proof. (1) Because AD is parallel to BC, and AC falls on them, therefore the angle DAC is equal to the alternate angle BCA,

Prop. 29 Again, because DC is parallel to AB, and AC falls on them, therefore the angle DCA is equal to the alternate angle BAC.

Because

(2) Because in the two triangles DAC, BAC

Prop. 29

Proved

Proved

the angle DAC is equal to the angle BCA, and the angle DCA is equal to the angle BAC, and the side AC is common to both triangles, therefore by Proposition 26 the triangles are equal in all respects. so that the side AD is equal to the side BC, and the side CD is equal to the side AB,

and the angle ADC is equal to the angle ABC,

and the triangle DAC is equal to the triangle BAC in area.

(3) Again, because the angle DAC is equal to the angle BCA, and the angle BAC is equal to the angle DCA,

therefore the whole angle BAD is equal to the whole angle BCD, Therefore the opposite sides and angles of the parallelogram ABCD are equal, and the diagonal AC bisects its area.

Q. E. D.