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

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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

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lines CE and AB, and, therefore, by the corollary to the last proposition, these lines are parallel.

PROP. XXVII. THEOR.

Parallel lines are equidistant, and equidistant straight lines are parallel.

The perpendiculars EG, FH, let fall from any points E, F in the straight line AB upon its parallel CD, are equal; and if these perpendiculars be equal, the straight lines AB and CD are parallel.

For join EH and because each of the interior angles EGH and FHG is a right angle, they are together equal to two right angles, and consequently the perpendiculars EG and FH are (Cor. I. 25.) pa- A E rallel to each other; wherefore

F

B

(I. 25.) the alternate angles HEG

and EHF are equal. But, EF be- C

H

D

ing parallel to GH, the alternate angles EHG and HEF are likewise equal; and thus the two triangles HGE and HFE, having the angles HEG and EHG respectively equal to EHF and HEF, and the side EH common to both, are (I. 23.) equal, and hence the side EG is equal to FH.

Again, if the perpendiculars EG and FH be equal, the two triangles EGH and EFH, having the side EG equal to FH, EH common, and the contained angle HEG equal to EHF, are (I. 3.) equal, and therefore the angle EHG equal to HEF, and (I. 25.) the straight line AB parallel to CD.

PROP. XXVIII. THEOR.

The opposite sides of a rhomboid are parallel.

If the opposite sides AB, DC, and AD, BC of the quadrilateral figure ABCD be equal, they are also parallel.

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For join AC. And because AB is equal to DC, BC to AD, and AC is common; the two triangles ABC and ADC are (I. 2.) equal. Consequently the angle ACD is equal to CAB, and the side AB D

(Cor. I. 25.) parallel to CD; and for the same reason, the angle CAD is equal to ACB, whence the side AD is parallel to BC.

Cor. Hence the angles of a square or rectangle are all of them right angles; for the opposite sides, being equal, are parallel; and if the angle at A be right, the other interior one at B is also a right angle (I. 25.), and consequently the angles at C and D, opposite to these, are right.

PROP. XXIX. THEOR.

The opposite sides and angles of a parallelogram are equal.

Let the quadrilateral figure ABCD have the sides AB, BC parallel to CD, AD; these are respectively equal, and so are the opposite angles at A and C, and at B and D.

For join AC. Because AB is parallel to CD, the alternate angles BAC and ACD are (I. 25.) equal; and since AD is parallel to BC, the alternate angles ACB and CAD are likewise equal. Wherefore the

D

C

triangles ABC and ADC, having the angles CAB and ACB equal to ACD and CAD, and the interjacent side AC common to both, are (I. 23.) equal. Consequently, the side AB is equal to CD, and the side BC to AD; and these opposite sides being thus equal, the opposite angles (I. 28.) must also be equal.,

Cor. Hence the diagonal divides a rhomboid or parallelogram into two equal triangles. Hence also an oblong

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 (1. 25.) equa!; 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. 3.) Whence the

side BC is equal to AD, and the angle

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B

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

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PROP. XXXI. THEOR.

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 (1. 25.), and likewise ABD and BDC. The tri- A angles AEB and CED, having thus the angles BAE, ABE respectively

equal to BCE and CED, and the in

B

E

D

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

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