angle at F be either obtuse or acute, the line EF, which forms it, can have only one corresponding position.Whence, in each of these three cases, the triangle ABC admits of a perfect adaptation with DEF.

PROP. XXV. THEOR.

If a straight line fall upon two parallel straight lines, it will make the alternate angles equal, the exterior angle equal to the interior opposite one, and the two interior angles on the same side together equal to two right angles.

Let the straight line EFG fall upon the parallels AB and CD; the alternate angles AGF and DFG are equal, the exterior angle EFC is equal to the interior angle EGA, and the interior angles CFG and AGF are together equal to two right angles.

For suppose the straight line EFG, produced both ways from F to turn about that point in the direction BA; it will first cut the extended line AB towards A, and will in its progress afterwards meet the same line on the other side towards B. In the position IFH, the angle EFH is the exterior angle of the triangle FHG, and therefore greater than FGH or EGA (I. 10.) But in the last position LFK, the exterior angle EFL is equal to its vertical angle GFK in the triangle FKG, and to which the angle FGA is exterior; consequently (I. 10.) FGA is greater than EFL, or the angle EFL is less than FGA or EGA. When the incident line EFG, therefore, meets AB above the point G, it makes an angle EFH greater than EGA; and when it meets AB below that point, it makes an angle EFL, which is less

B

K

G

H

I

than the same angle. But in passing through all the de

grees from greater to less, a varying magnitude must evidently rencounter, as it proceeds, the single intermediate limit of equality. Wherefore, there is a certain position, CD, in which the line revolving about the point F makes the exterior angle EFC equal to the interior EGA, and at the same time meets AB neither towards the one part nor the other, or is parallel to it.

And now, since EFC is proved to be equal to EGA, and is also equal to the vertical angle GFD; the alternate angles FGA and GFD are equal. Again, because GFD and FGA are equal, add the angle FGB to each, and the two angles GFD and FGB are equal to FGA and FGB; but the angles FGA and FGB, on the same side of AB, are equal to two right angles, and consequently the interior angles GFD and FGB are likewise equal to two right angles.

Cor. Since the position CD is individual, or that only one straight line can be drawn through the point F parallel to AB, it follows that the converse of the proposition is true, and that those three properties of parallel lines are also the criteria for distinguishing parallels.

PROP. XXVI. PROB.

Through a given point, to draw a straight line parallel to a given straight line.

To draw, through the point C, a straight line parallel to AB.

In AB take any point D, join CD, and at the point C make (I. 4.) an angle DCE equal to CDA; CE is parallel to AB.

For the angles CDA and DCE, thus formed equal, are the alternate angles which CD makes with the straight

is a rectangular parallelogram; for if the angle at A be right, the opposite angle at C is right, and the remaining angles at B and D, being equal to each other and to two right angles, must be right angled.

PROP. XXX. THEOR.

If the parallel sides of a trapezoid be equal, the other sides are likewise equal and parallel.

Let the sides AB and DC be equal and parallel; the sides AD and BC are themselves equal and parallel.

For join AC. Because AB is parallel to CD, the alternate angles CAB and ACD are (I. 25.) equal; and the triangles ABC and ADC, having the side AB equal to CD, AC common to both, and the contained angle CAB equal to ACD, are, therefore, equal (I. S.) Whence the side BC is equal to AD, and the angle

B

ACB equal to CAD; but these angles being alternate, BC must also be parallel to AD (Cor. I. 25.)

The diagonals of a rhomboid mutually bisect each other.

If the diagonals of the rhomboid ABCD intersect each other in E; the part AE is equal to CE, and DE to BE. For because a rhomboid is also a parallelogram (I. 28.), the alternate angles BAC and ACD are equal (I. 25.), and likewise ABD and BDC. The tri- A angles AEB and CED, having thus the angles BAE, ABE respectively

equal to DCE and CED, and the in

B

E

D

terjacent sides AB and CD equal, are (1. 23.) wholly equal.

Wherefore AE is equal to the corresponding side CE, and BE to DE.

Cor. Hence the diagonals of a rectangle are equal to each other; for if the angles at A and B were right angles, the triangles DAB and CBA would be equal (I. 3.) and consequently the base DB equal to AC.

PROP. XXXII. THEOR.

Lines parallel to the same straight line, are parallel to each other.

If the straight line AB be parallel to CD, and CD parallel to EF; then is AB parallel to EF.

For let the straight line GH cut these lines.

A I

B

K

E

LF

C

Because AB is parallel to CD, the exterior angle GIA is equal (I. 25.) to the interior GKC; and since CD is parallel to EF, this angle GKC is, for the same reason, equal to GLE. Therefore GIA is equal to GLE, and consequently AB is parallel to EF (I. 25. Cor.)

PROP XXXIII. THEOR. (

H

Straight lines drawn parallel to the sides of an angle, contain an equal angle.

If the straight lines AB, AC be parallel to DE, DF; the angle BAC is equal to EDF.

For draw the straight line GAD through the vertices. And because AC is paral- D4 lel to DF, the exterior angle GAC is (I. 25.) equal to GDF; and for the same

B

A

I

F

reason, GAB is equal to GDE; there consequently remains the angle BAC equal to EDF.

C

PROP. XXXIV. THEOR.

An exterior angle of a triangle is equal to both its opposite interior angles, and all the interior angles of a triangle are together equal to two right angles.

The exterior angle BCD, formed by the production of the side AC of the triangle ABC, is equal to the two opposite interior angles CAB and CBA, and all the interior angles CAB, CBA, and BCA of the triangle, are together equal to two right angles.

For through the point C draw (I. 26.) the straight line CE parallel to AB. And because AB is parallel to CE, the interior angle BAC is (I. 25.) equal to the exterior one ECD; and for the same reason the alternate angle ABC is equal to BCE. Wherefore the two angles CAB and ABC are equal to DCE and ECB, or to

B

E

the whole exterior angle BCD. Add the adjacent angle BCA to each; and all the

interior angles of the triangle ABC are

together equal to the angles BCD and A

D

BCA on the same side of the straight line AD, that is, to two right angles.

Cor. 1. Hence the two acute angles of a right angled triangle are together equal to one right angle; and hence each angle of an equilateral triangle is two third parts of a right angle.

Cor. 2. Hence if a triangle have its exterior angle, and one of its opposite interior angles, double of those of another triangle; its remaining opposite interior angle will also be double of the corresponding angle of the other.