REMARKS.

It must be observed that two parallels, and a straight line intersecting them, are a special case of the triangle, the vertex, or intersection of two of the lines, being removed to an infinite distance. In Th. 9 it was proved that the exterior angle of a triangle is greater than the interior and opposite angle: from which the contra-positive theorem (Th. 21) logically follows, that if the exterior angle is equal to the interior and opposite angle, the lines do not form a triangle.

Theorem 22 is in fact proved by the rule of identity (p. 4).

Since there is only one straight line through B that makes the alternate angles equal;

and only one straight line through B that is parallel to GH;

(Ax. 5.)

and the line that makes the alternate angles equal is the parallel;

(Th. 21.)

therefore the parallel makes the alternate angles equal.

THEOREM 25.

If one side of a triangle be produced the exterior angle will be equal to the two interior and opposite angles, and the three interior angles of a triangle are together equal to two right angles.

Let one side BC of the triangle ABC be produced to D: then shall the angle ACD = the sum of the angles ABC, CAB; and the three angles ABC, BCA, CAB shall be together equal to two right angles.

Proof. For if through C a line CH were drawn parallel to BA,

the angle HCD= the corresponding angle ABC, (Th. 22.) and the angle ACH the alternate angle BAC;

=

(Th. 22.)

.: the whole angle ACD = the two angles ABC+BAC.

Again, if ACB be added to these,

the two angles ACD + ACB = the three angles ABC+ ВСА + САВ.

But ACD + ACB = two right angles;

(Th. 4.)

therefore ABC + BCA + CAB = two right angles*. Q. E. D. COR. In a right-angled triangle the two acute angles together make up one right angle.

THEOREM 26.

The interior angles of any polygon are together less than twice as many right angles as the figure has sides by four right angles.

Part. En. Let ABCDE be any polygon; it is required to prove that its interior angles are together less than twice as many right angles as the figure has sides by four right angles.

Proof. Take any point O within the polygon; and join OA, OB, OC, OD, OE.

Then there are as many triangles having O as a common vertex as the figure has sides.

And, since the interior angles of a triangle are equal to

two right angles;

(Th. 25.)

* Euclid, I. 32.

therefore all the angles of all the triangles are equal to twice as many right angles as the figure has sides.

But all the angles of all the triangles make up all the angles of the polygon together with the angles at O, which are equal to four right angles; (Th. 4. Cor.)

therefore all the angles of the polygon, together with four right angles, are equal to twice as many right angles as the figure has sides;

that is, all the angles of the polygon are together less than twice as many right angles as the figure has sides by four right angles*.

COR. The exterior angles of any convex polygon are together equal to four right angles.

Let ABCDE be a convex polygon having all its sides AB, BC, CD, DE, EA produced;

it is required to prove that the sum of its exterior angles is equal to four right angles.

Proof. Each interior angle together with its adjacent exterior angle are equal to two right angles;

(Th. 2.) therefore all the interior angles together with all the exterior angles are equal to twice as many right angles as the figure has sides;

* Euclid, 1. 32. Cor. I.

but all the interior angles, together with four right angles, are equal to twice as many right angles as the figure has sides;

(Th. 26.)

therefore all the exterior angles are equal to four right angles*.

Q. E. D.

THEOREM 27.

The adjoining angles of a parallelogram are supplementary and the opposite angles are equal.

Part. En. Let HBGE be a parallelogram;

that is, let HE, EG be respectively parallel to BG, BH;

(Def. 37-)

it is required to prove that its adjoining angles EHB, HBG are supplementary, and its opposite angles HBG, HEG are equal.

Proof. Because HE is parallel to BG, and HB meets them,

(Hyp.)

therefore HBG is supplementary to EHB.

(Th. 23. Cor.)

(Hyp.)

And because HB is parallel to EG,

and HE meets them,

therefore HEG is supplementary to EHB; (Th. 23. Cor.)

but HBG is also supplementary to EHB,

therefore HEG is equal to HBG.

*

(Th. 1. Cor. 3.) Q. E. D.

COR. I. Hence if one of the angles of a parallelogram is a right angle, all its angles are right angles.

COR. 2. If two straight lines are respectively parallel to two other straight lines they will include equal angles towards the same parts.

Def. 40. A right-angled parallelogram is called a rectangle.

THEOREM 28.

The opposite sides of a parallelogram are equal to one another, and the diagonal bisects it.

Part. En. Let ABCD be a parallelogram, that is, let AB be parallel to CD, and AD to BC; it is required to prove that AB is equal to DC, and AD to BC.

Proof. Join AC.

Then because AB is parallel to DC, and AC meets them;

(Hyp.)

therefore the angle BAC is equal to the alternate angle ACD.

(Th. 22.)

(Hyp.)

And because AD is parallel to BC; therefore the angle BCA is equal to the alternate angle

CAD:

therefore in the triangles BAC, DCA we have

the angle BAC = the angle DCA,

and the angle BCA = the angle DAC;

and the side AC adjacent to the equal

angles common;

(Th. 22.)

therefore the triangles are equal in all respects, (Th. 7.)