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THEOREM XIV.

When a line cuts two parallel lines, the outward angle is equal to the inward opposite one, on the same side; and the two inward angles, on the same side, are toyether equal to two right angles.

LET the line EF cut the two parallel lines AB, CD; then will the outward angle EGB be equal to the inward opposite angle GHD, on the same side of the line EF; and the two inward angles BGH, GHD, taken together, will be equal to two right angles.

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For since the two lines AB, CD, are parallel, the angle AGH is equal to the alternate angle GHD (th. 12). But the angle AGH is equal to the opposite angle EGB (th. 5.) Therefore the angle EGB is also equal to the angle GHD (ax. 1).

Again, because the two adjacent angles EGB, BGH, are together equal to two right angles (th. 4), of which the angle EGB has been shown to be equal to the angle GHD; therefore the two angles BGH, GHD, taken together, are also equal to two right angles.

Cor. 1. And, conversely, if one line meeting two other lines, make the angles on the same side of it equal, those two lines are parallels.

Cor. 2. If a line, cutting two other lines, make the sum of the two inward angles on the same side, less than two right angles, those two lines will not be parallel, but will meet each other when produced.

THEOREM XV.

Those lines which are parallel to the same line are parallel to each other.

LET the lines AB, CD, be each of them parallel to the line EF; then shall the lines AB, CD, be parallel to each other. For, let the line GI be perpendicular to EF. Then will this line be also perpendicular to both the lines AB, CD (cor. th. 12), and consequently the two lines AB, CD, are parallels (cor. th. 13).

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THEOREM XVI.

When one side of a triangle is produced, the outward angle is equal to both the inward opposite angles taken together.

LET the side AB, of the triangle ABC, be produced to D; then will the outward angle CBD be equal to the sum of the two inward opposite angles A and C.

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For, conceive BE to be drawn parallel to the side AC of the triangle. Then BC, meeting the two parallels AC, BE, makes the alternate angles C and CBE equal (th. 12). And AD, cutting the same two parallels AC, BE, makes the inward and outward angles on the same side, A and EBD, equal to each other (th. 14). Therefore, by equal additions, the sum of the two angles A and C, is equal to the sum of the two CBE and EBD, that is, to the whole angle CBD (ax. 2).

THEOREM XVII.

In any triangle, the sum of all the three angles is equal to two right angles. LET ABC be any plane triangle; then the sum of the three angles, A + B + C, is equal to two right angles.

For, let the side AB be produced to D. Then the outward angle CBD is equal to the sum of the two inward opposite angles A+ C (th. 16). To each of these equals A add the inward angle B, then will the sum of the three

inward angles, A + B + C, be equal to the sum of the two adjacent angles ABC + CBD (ax. 2). But the sum of these two last adjacent angles is equal to two right angles (th. 4). Therefore also the sum of the three angles of the triangle, A + B + C, is equal to two right angles (ax. 1).

Cor. 1. If two angles in one triangle be equal to two angles in another triangle, the third angles will also be equal (ax. 3), and the two triangles equiangular.

Cor. 2. If one angle in one triangle, be equal to one angle in another, the sums of the remaining angles will also be equal (ax. 3).

Cor. 3. If one angle of a triangle be right, the sum of the other two will also be equal to a right angle, and each of them singly will be acute, or less than a right angle.

Cor. 4. The two least angles of every triangle are acute, or each less than a right angle.

THEOREM XVIII.

In any quadrangle, the sum of all the four inward angles is equal to four right angles.

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LET ABCD be a quadrangle; then the sum of the four inward angles, A+B+C+D, is equal to four right angles. Let the diagonal AC be drawn, dividing the quadrangle into two triangles, ABC, ADC. Then, because the sum of the three angles of each of these triangles is equal to two right angles (th. 17); it follows, that the sum of all the angles of both triangles, which make up the four angles of the quadrangle, must be equal to four right angles (ax. 2).

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Cor. 1. Hence, if three of the angles be right ones, the fourth will also be a right angle.

Cor. 2. And if the sum of two of the four angles be equal to two right angles, the sum of the remaining two will also be equal to two right angles.

THEOREM XIX.

In any figure whatever, the sum of all the inward angles, taken together, is equal to twice as many right angles, wanting four, as the figure has sides.

LET ABCDE be any figure; then the sum of all its inward angles, A + B + C + D + E, is equal to twice as many right angles, wanting four, as the figure has sides.

For, from any point P, within it, draw lines, PA, PB, PC, &c. to all the angles, dividing the polygon into as many triangles as it has sides. Now the sum of the three angles of each of these triangles, is equal to two right angles (th. 17);

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therefore the sum of the angles of all the triangles is equal to twice as many right angles as the figure has sides. But the sum of all the angles about the point P, which are so many of the angles of the triangles, but no part of the inward angles of the polygon, is equal to four right angles, (cor. 3, th. 4,) and must be deducted out of the former sum. Hence it follows that the sum of all the inward angles of the polygon alone, A + B + C + D + E, is equal to twice as many right angles as the figure has sides, wanting the said four right angles.

THEOREM XX.

When every side of any figure is produced, the sum of all the outward angles thereby made is equal to four right angles.

LET A, B, C,.... be the outward angles of any polygon, made by producing all the sides; then will the sum, A + B + C + D + E, of all those outward angles, be equal to four right angles.

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For every one of these outward angles, together with its adjacent inward angle, make up two right angles, as A+ a equal to two right angles, being the two angles made by one line meeting another (th. 4). And there being as many outward, or inward angles, as the figure has sides; therefore the sum of all the inward and outward angles, is equal to twice as many right angles as the figure has sides. But the sum of all the inward angles, with four right angles, is equal to twice as many right angles as the figure has sides (th. 19). Therefore the sum of all the inward and all the outward angles is equal to the sum of all the inward angles and four right angles (ax. 1). From each of these take away all the inward angles, and there remain all the outward angles equal to four right angles (by ax. 3).

THEOREM XXI.

A perpendicular is the shortest line that can be drawn from a given point to an indefinite line: and, of any other lines drawn from the same point, those that are nearest the perpendicular are less than those more remote.

IF AB, AC, AD,.... be lines drawn from the given point A, to the indefinite line DE, of which AB is perpendicular; then shall the perpendicular AB be less than AC, and AC less than AD, &c.

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For, the angle B being a right one, the angle C is acute (cor. 3, th. 17), and therefore less than the angle B. But the less angle of a triangle is subtended by the less side (th. 9). Therefore the side AB is less than the side AC.

Again, the angle ACB being acute, as before, the adjacent angle ACD will be obtuse (th. 4); consequently the angle D is acute (cor. 3, th. 17), and therefore is less than the angle C. And since the less side is opposite to the less angle, therefore the side AC is less than the side AD.

Cor. A perpendicular is the least distance of a given point from a line.

THEOREM XXII.

The opposite sides and angles of any parallelogram are equal to each other; and the diagonal divides it into two equal triangles.

LET ABCD be a parallelogram, of which the diagonal is BD; then will its opposite sides and angles be equal to each other, and the diagonal BD will divide it into two equal parts, or triangles.

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For, since the sides AB and DC are parallel, as also the sides AD and BC (def. 37), and the line BD meets them; therefore the alternate angles are equal (th. 12), namely, the angle ABD to the angle CDB, and the angle ADB to the angle CBD: hence the two triangles, having two angles in the one equal to two angles in the other, have also their third angles equal (cor. 1, th. 17), namely, the angle A equal to the angle C, which are two of the opposite angles of the parallelogram.

Also, if to the equal angles ABD, CDB, be added the equal angles CBD, ADB, the wholes will be equal (ax. 2), namely, the whole angle ABC to the whole ADC, which are the other two opposite angles of the parallelogram.

Again, since the two triangles are mutually equiangular and have a side in each equal, viz. the common side BD; therefore the two triangles are identical (th. 2), or equal in all respects, namely, the side AB equal to the opposite side DC, and AD equal to the opposite side BC, and the whole triangle ABD equal to the whole triangle BCD,

Cor. 1. Hence, if one angle of a parallelogram be a right angle, all the other three will also be right angles, and the parallelogram a rectangle.

Cor. 2. Hence also, the sum of any two adjacent angles of a parallelogram is equal to two right angles.

THEOREM XXIII.

Every quadrilateral, whose opposite sides are equal, is a parallelogram, or has its opposite sides parallel.

LET ABCD be a quadrangle, having the opposite sides equal, namely, the side AB equal to DC, and AD equal to BC; then shall these equal sides be also parallel, and the figure a parallelogram.

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For, let the diagonal BD be drawn. Then, the triangles, ABD, CBD, being mutually equilateral (hyp.), they are also mutually equiangular (th. 8), or have their corresponding angles equal; consequently the opposite sides are parallel (th. 13); viz. the side AB parallel to DC, and AD parallel to BC, and the figure is a parallelogram.

THEOREM XXIV.

Those lines which join the corresponding extremes of two equal and parallel lines, are themselves equal and parallel.

LET AB, DC, be two equal and parallel lines; then will the lines AD, BC, which join their extremes, be also equal and parallel. (See the fig. above.) For, draw the diagonal BD. Then, because AB and DC are parallel, (hyp.)

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the angle ABD is equal to the alternate angle BDC (th. 12): hence then, the two triangles having two sides and the contained angles equal, viz. the side AB equal to the side DC, and the side BD common, and the contained angle ABD equal to the contained angle BDC, they have the remaining sides and angles also respectively equal (th. 1); consequently AD is equal to BC, and also parallel to it (th. 12).

THEOREM XXV.

Parallelograms, as also triangles, standing on the same base, and between the same parallels, are equal to each other.

LET ABCD, ABEF, be two parallelograms, and ABC, ABF, two triangles, standing on the same base AB, and between the same parallels AB, DE; then will the parallelogram ABCD be equal to the parallelogram ABEF, and the triangle ABC equal to the triangle ABF.

For, since the line DE cuts the two parallels AF, BE, and the two AD, BC, it makes the angle E equal to the angle AFD, and the angle D equal to the angle BCE

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(th. 14); the two triangles ADF, BCE, are therefore equiangular (cor. 1, th. 17); and having the two corresponding sides AD, BC, equal (th. 22), being opposite sides of a parallelogram, these two triangles are identical, or equal in all respects (th. 2). If each of these equal triangles then be taken from the whole space ABED, there will remain the parallelogram ABEF in the one case, equal to the parallelogram ABCD in the other (ax. 3).

Also the triangles ABC, ABF, on the same base AB, and between the same parallels, are equal, being the halves of the said equal parallelograms (th. 22). Cor. 1. Parallelograms, or triangles, having the same base and altitude, are equal. For the altitude is the same as the perpendicular or distance between the two parallels, which is every where equal, by the definition of parallels.

Cor. 2. Parallelograms, or triangles, having equal bases and altitudes, are equal. For, if the one figure be applied with its base on the other, the bases will coincide or be the same, because they are equal: and so the two figures, having the same base and altitude, are equal.

THEOREM XXVI.

If a parallelogram and a triangle stand on the same base, and between the same parallels, the parallelogram will be double the triangle, or the triangle half the parallelogram.

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LET ABCD be a parallelogram, and ABE a triangle, on the same base AB, and between the same parallels AB, DE; then will the parallelogram ABCD be double the triangle ABE, or the triangle half the parallelogram. For, draw the diagonal AC of the parallelogram, dividing it into two equal parts (th. 22). Then because the triangles ABC, ABE, on the same base, and between the same parallels, are equal (th. 25), and because the one triangle ABC is half the parallelogram ABCD (th. 22), the other equal triangle ABE is also equal to half the same parallelogram ABCD.

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