greater angle (th. 9), the side CD (of the triangle BCD) is less than the side BC. Q. E. D.

THEOREM XII.

WHEN a Line Intersects two Parallel Lines, it makes the Alternate Angles Equal to each other.

Let the line EF cut the two parallel lines AB, CD; then will the angle AF be equal to the alternate augle aFD.

A

E B

C

D

For if they are not equal, one of them must be greater than the other; let it be EFD for instance which is the greater if possible; and conceive the line FB to be drawn; cutting off the part or angle EFB equal to the angle AF; and meeting the line AB in the point B.

Then, since the outward angle AEF, of the triangle BEF, is greater than the inward opposite angle EFB (th. 8); and since these two angles also are equal (by the constr.) it follows, that those angles are both equal and unequal at the same time: which is impossible. Therefore the angle EFD is not unequal to the alternate angle AEF, that is, they are equal to each other. Q. E. D.

Corol. Right lines which are perpendicular to one, of two parallel lines, are also perpendicular to the other.

THEOREM XIII.

WHEN a line, cutting Two other Lines, makes the Al ternate Angles Equal to each other, those two Lines are Parallel.

Let the line EF, cutting the two lines AB, CD, make the alternate angles AEF, DFE, equal to each other; then will AB be parallel to CD.

For if they be not parallel, let some other line, as FG, be parallel to AB. Then, because of these parallels, the

A

B

D

G

But

angle AEF is equal to the alternate angle EFG (th. 12). the angle AEF is equal to the angle EFD (by hyp.). Therefore the angle EFD is equal to the angle EFG (ax. 1); that is, a part is equal to the whole, which is impossible. Therefore no line but CD can be parallel to AB.

Q. E. D.

Corol. Those lines which are perpendicular to the same line, aré parallel to each other.

THEOREM

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, equal to two Right Angles.

Let the line EF cut the two parallel lines AB, CD; then will the outward angle A EGB be equal to the inward opposite angle GHD, on the same side of the line EF;

and the two inward angles BGH, GиD, C taken together, will be equal to two right angles.

F

G

E

B

D

H

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. 7). Therefore the angle EGB is also equal to the angle GHD (ax. 1). Q. E. D.

Again, because the two adjacent angles EGB, BGH, are together equal to two right angles (th. 6); of which the angle EGB has been shown to be equal to the angie GHD; therefore the two angles BGH, GHD, taken together, are also equal to two right angles.

Corol. 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.

Corol. 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, afe Parallel to each other.

E

-F

I

For, let the line GI be perpendicular to Er. Then will this line be also perpendicular to both the lines AB, CD, (corol. th. 12), and con sequently the two lines AB, CD, are parallels (corol. th, 13).

2. E. D.

VOL. I.

38

THEOREM

THEOREM XVI.

WHEN One Side of a triangle is produced, the Outward Angle is equal to both the Inward Opposite Angles taken together.

1

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

A

B

D

Then BC, meeting the

For, conceive BE to be drawn parallel to the side AC of the triangle. 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 (by ax. 2). Q. E. D.

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.

C

A

B D

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 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. 6). Therefore also the sum of the three angles of the triangle A+B+c is equal to two right angles (ax. 1). Q. E. d.

Corol. 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.

Corol. 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).

Corol

Corol. 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.

Corol. 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.

Let ABCD be a quadradrangle; then the sum of the four inward angles, A+B+C+ D is equal to four right angles.

A

D

B

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). Q. E. D.

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

Corol. 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.

E

P

B

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); 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

many of the angles of the triangles, but no part of the inward angles of the polygon, is equal to four right angles (corol. 3, th. 6), 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.* Q. E. D.

THEOREM XX.

WHEN every Side of any Figure is produced out, the Sum of all the Outward Angles thereby made, is equal to Four Right Angles.

Let A, B, C, &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.

E

B

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. 6). 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 (by 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, &c. 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.

For the angle в being a right one, the

A

DCBE

* This demonstration does not apply to all rectilineal figures. Ed.

angle