PROPOSITION XVII. THEOREM.

If from C, the middle point of the straight line AB, the line EF be drawn perpendicular to this straight line, then (first) every point of the perpendicular will be equally distant from the two extremities of AB; and (secondly) every point situated without the perpendicular will be unequally distant from those extremities.

First. Since we suppose AC-CB, the two oblique lines AD, DB, are equally distant from the perpendicular, and therefore equal. So, likewise, are the two oblique lines AE, EB, the two AF, FB, and so on. Therefore every point in the perpendicular is equally distant from the extremities A and B.

Secondly. Let I be a point out of the perpendicular. If IA and IB be joined, one of those lines will cut the perpendicular in D;

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from which drawing DB, we shall have DB-DA. But the straight line IB is less than ID+DB, and ID+DB=ID+ DA=IA; therefore IBIA; therefore every point out of the perpendicular is unequally distant from the extremities A and B.

PROPOSITION XVIII. THEOREM.

Two right-angled triangles are equal, when the hypotenuse and a side of the one are respectively equal to the hypotenuse and a side of the other.

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Their equality would be manifest, if the third sides BC and EF were equal. If possible, suppose that those sides are not equal, and that BC is the greater. Take BG EF; and join AG. The triangle ABG is equal to DEF; for the right angles B and E are equal, the side ABD E, and BG= EF; hence these triangles are equal (Prop. 6.), and consequently AG=DF. Now (Hyp.) we have DF-AC; and there

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fore AG=AC. But (Prop. 16.) the oblique line AC cannot be equal to AG, which lies nearer the perpendicular AB; therefore it is impossible that BC can differ from EF; therefore the triangles ABC and DEF are equal.

PROPOSITION XIX. THEOREM.

If two straight lines AC, BD, (see the next figure), are perpendicular to a third AB, they will be parallel, in other words (Def. 12.), they will never meet, how far soever both of them be produced.

For if they could meet in a point O, on either side of AB, there would be two perpendiculars OA, OB, let fall from the same point, on the same straight line, which is impossible.

PROPOSITION XX. LEMMA.

If the straight line BD is perpendicular to AB, and if another straight line AE makes with AB the acute angle BAE, then the straight lines BD, AE, if produced sufficiently, will meet.

Now, let another point L be taken in the line AE, at a distance AL greater than AF, and let LM be drawn from it perpendicular to AB. We can prove, as in the preceding case, that the point M cannot fall on G, or in the direction GA; it must fall therefore in the direction OB; so that the distance AM will, of necessity, be greater than AG.

I observe farther, that if the figure be constructed with care, and AL be taken double of AF, we shall find that AM is exactly double of AG; in like manner, if AL is taken triple of

AF, we shall find that AM is triple of AG; and, in general, that there is always the same proportion between AM and AG, as between AL and AF. This proportion being settled, it follows not only that the straight line AE, if produced sufficiently, will meet BD, but also that the distance on AE, of this point of concourse, may be accurately assigned. It will be the fourth term of the proportion, AG: AB::AF: x.

Scholium. The preceding investigation, being founded on a property which is not deduced from reasoning alone, but discovered by measurements made on a figure constructed accurately, has not the same character of rigorousness with the other demonstrations of elementary geometry. It is given here merely as a simple method of arriving at a conviction of the truth of the proposition. For a strictly rigorous demonstration we refer to the second Note.

PROPOSITION XXI. THEOREM.

If two straight lines, AC, BD, form with a third AB, two interior angles CAB, ABD, whose sum is equal to two right angles, the lines AC, BD, will be parallel.

From G, the middle point of AB, draw the straight line EGF perpendicular to AC.

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will also be perpendicular to BD. For the sum GAE+GBD is (Hyp.) equal to two right angles; the sum GBF+GBD is (Prop. 2.) likewise equal to two right angles; and taking away GBD from both, there remains the angle GAE GBF. Again, the angles AGE, BGFareequal (Prop.5.), therefore the triangles AGE and BGF have each

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a side and two adjacent angles equal; therefore (Prop. 7.) they are themselves equal, and the angle BFG is equal to AEG: but AEG is a right angle by construction; therefore the line AC, BD, being perpendicular to the same straight line EF, are parallel (Prop. 19.).

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PROPOSITION XXII. THEOREM.

If two straight lines AI, BD, make with a third AB, two interior angles BAI, ABD, whose sum is less than two right angles, the lines AI, BD will meet if produced.

Draw the straight line AC (see the preceding figure), making the angle CABABF, in other words, so that the angles CAB ABD, taken together, may be equal to two right angles; and complete the rest of the construction, as in the foregoing theorem. Since AEK is a right angle, AE the perpendicular is shorter than AK the oblique line; hence (Prop. 14.) in the triangle AEK, the angle AKE, opposite the side AE, is less than the right angle AEK, opposite the side AK. Hence the angle IKF, equal to AKE, is less than a right angle; hence (Prop. 20.) the lines KI and FD will meet if produced.

Scholium. If the lines AM, BD make with AB two angles BAM, ABD, the sum of which is greater than two right angles, those lines will not meet above AB, but they will below. For the two angles BAM, BAN are together equal to two right angles, and so are the two ABD, ABF; hence those four angles are together equal to four right angles. But BAM, ABD are together greater than two right angles; therefore the remaining angles BAN, ABF are together less than two; therefore the straight lines AN, BF will meet if produced.

Cor. Through a given point A, no more than one line can be drawn parallel to a given line BD. For there is but one line AC, which makes the sum of the two angles BAC+ABD equal to two right angles; and this is the parallel required. Every other line AI or AM would make the sum of the interior angles less or greater than two right angles; therefore it would meet BD.

PROPOSITION XXIII. THEOREM.

If two parallels AB, CD are met by a secant EF, the sum of the interior angles AGO, GOC will be equal to two right angles.

For if it were more or less, the two straight lines AB, CD would meet on the one side or the other (Prop. 22.), and would not be rallel.

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Cor. 1. If GOC is a right angle, AGO will be a right angle also; therefore every line perpen

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dicular to one of two parallels is perpendicular to the other.

Cor. 2. Since the sum AGO+GOC is equal to two right angles, and the sum GOD+GOC is also equal to two, if GOC be taken from both, there will remain the angle AGO GOD. Besides (Prop. 5.), we have AGO-BGE, and GOD COF; hence the four acute angles AGO, BGE, GOD, COF are equal to each other. The same is the case with the four obtuse angles AGE, BGO, GOC, DOF. It may be observed, moreover, that, in adding one of the acute angles to one of the obtuse, the sum will always be two right angles.

Scholium. The angles just spoken of, when compared with each other, assume different names. AGO, GOC, we have already named interior angles on the same side; BGO, GOD have the same designation; AGO, GOD are called alternate interior angles, or simply alternate; so also are BGO, GOC: and lastly EGB, GOD, or EGA, GOC are called, respectively, the opposite exterior and interior angles; and EGB, COF, or AGE, DOF the alternate exterior angles. This being premised, the following propositions may be considered as already demonstrated.

First. The interior angles on the same side are together equal to two right angles.

Second. The alternate interior angles are equal; so likewise are the opposite exterior and interior, and the alternate exterior angles.

Conversely, if in this second case, two angles of the same name are equal, the lines to which they refer will be parallel. Suppose, for example, the angle AGO GOD. Since GOC+ GOD is equal to two right angles, AGO+GOC must also be equal to two, and (Prop. 21.) the lines AG, CO must be parallel.

PROPOSITION XXIV. THEOREM.

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Two lines AB, CD, parallel to a third, are parallel to each other. Draw PQR perpendicular to EF, and cutting AB, CD. Since AB is parallel to EF, PR will be perpendicular to AB (Prop. 23. Cor. 1.); and since CD is parallel to EF, PR will for a like reason be perpendicular to CD. Hence AB and CD are perpendicular to the same straight line; hence (Prop. 19.) they are parallel.

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