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F

points, E and F, in one of them,

G draw the lines EG, FH perpendic- C

D ular to AB; they will also be perpendicular to CD (Prop. XXIII., Cor. 1). Join EH; then, because

A

E B EG and FH are perpendicular to the same straight line AB they are parallel (Prop. XX.); therefore, the alternate angles, EHF, HEG, which they make with HE are equal (Prop. XXIII.). Again, because AB is parallel to CD, the alternate angles GHE, HEF are also equal. Therefore, the triangles HEF, EHG have two angles of the one equal to two angles of the other, each to each, and the side Ell included between the equal angles, common; hence the triangles are equal (Prop. VII.); and the line EG, which measures the distance of the parallels at the point E, is equal tu the line FH, which measures the distance of the same parallels at the point F. Therefore, two parallel straight lines, &c.

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Two angles are equal, when their sides are purallel, each to each, and are similarly situated. Let BAC, DEF be two angles, having

B che side BA parallel to DE, and AC to EF; the two angles are equal to each

E other.

Produce DE, if necessary, until it meets A AC in G. Then, because EF is parallel to GC, the angle DEF is equal to DGC

Н. E

F (Prop. XXIII.); and because DG is parallel to AB, the angle DGC is equal to BAC; hence the angle DEF is equal to the angle BAC (Axiom 1). Therefore, two angles, &c.

Scholium. This proposition is restricted to the case in which the sides which contain the angles are similarly situated; because, if we produce FE to H, the angle DEH has its sides parallel to those of the angle BAC; but the two angles are not equal.

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If one side of a triangle is produced, the exterior angle as equal to the sum of the two interior and opposite angles; and the three interior angles of every triangle are equal to two right angles.

Let ABC be any plane triangle, and let the side BC be

A

B

produced 10 D; then will the exterior angle ACD be equal to the

E sum of the two interior and opposite angles A and B; and the sum of the three angles ABC, BCA, CAB is equal to two right angles.

с

D For, conceive CE to be drawn parallel to the side AB of the triangle ; then, because AB is parallel to CE, and AC meets them, the alternate angles BAC, ACE are equal (Prop. XXIII.). Again, because AB is parallel to CE, and BD meets them, the exterior angle ECD is equal to the interior and opposite angle ABC. But the angle ACE was proved equal to BAC; therefore the whole exterior' angle ACD is equal to the two interior and opposite angles CAB, ABC (Axiom 2). To each of these equals add the angle ACB; then will the sum of the two angles ACD, ACB be equal to the sum of the three angles ABC, BCA, CAB. But the angles ACD, ACB are equal to two right angles (Prop. II.) ; hence, also, the angles ABC, BCA, CĂB are together equal to two right angles. Therefore, if one side of a triangle, &c.

Cor. 1. If the sum of two angles of a triangle is given, the third may be found by subtracting this sum from two right angles.

Cor. 2. If two angles of one triangle are equal to two angles of another triangle, the third angles are equal, and the triangles are mutually equiangular.

Cor. 3. A triangle can have but one right angle; for if there were two, the third angle would be nothing. Still less can a triangle have more than one obtuse angle.

Cor. 4. In a right-angled triangle, the sum of the two acute angles is equal to one right angle.

Cor. 5. In an equilateral triangle, each of the angles is one third of two right angles, or two thirds of one right angle.

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The sum of all the interior angles of a polygon, is equal to twice as many right angles, wanting four, as the figure has sides.

Let ABCDE

be any polygon; then the sum of all its interior angles A, B, C, D, E is equal to twice as many right an. gles, wanting four, as the figure has sides (see next page).

For, from any point, F, within it, draw lines FA, FB, FC, &c, to all the angles. The polygon is thus divided into as many tri ingles as it has sides. Now the sum of the three

D

E

А

angles of each of these triangles, is equal to two right angles (Prop. XXVII.); therefore the sum of the angles of all the triangles, is equal to twice as many right angles as the polygon has sides. But the same angles are equal to the angles of the polygon, together with the angles at the point F, that is, together with four

B right angles (Prop. V., Cor. 2). Therefore the angles of the polygon are equal to twice as many right angles as the figure has sides, wanting four right angles.

Cor. 1. The sum of the angles of a quadrilateral is four right angles ; of a pentagon, six right angles; of a hexagon, eight, &c.

Cor. 2. All the exterior angles of a polygon are together equal to four right angles. Because every interior angle, ABC, together with its adjacent exterior angle, ABD, is equal to two right angles (Prop. II.); therefore the sum of all the interior and exterior angles, is equal to twice as many right angles as the polygon has sides; that is, they are equal to all the interior angles of the polygon, together with four right angles. Hence Ō the sum of the exterior angles must be equal to four right angles (Axiom 3).

IC

B

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The opposite sides and angles of a parallelogram are equal to each other.

D

Let ABDC be a parallelogram ; then will

A

B its opposite sides and angles be equal to each other.

Draw the diagonal BC; then, because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal to each other (Prop. XXIII.). Also, because AC is parallel to BD, and BC meets them, the alternate angles BCA, CBD are equal to each oth

Hence the two triangles ABC, BCD have two angles, ABC, BCA of the one, equal to two angles, BCD, CBD, of the other, each to each, and the side BC included between chese equal angles, common to the two triangles; therefore their other sides are equal, each to each, and the third angle of the one to the third angle of the other (Prop. VII.), viz.

er.

the side AB to the side CD, and AC to BD, and the angle BAC equal to the angle BDC. Also, because the angle ABC is equal to the angle BCD, and the angle CBD to the angle BCA, the whole angle ABD is equal to the whole angle ACD. But the angle BAC has been proved equal to the angle BDC; therefore the opposite sides and angles of a parallelogram are equal to each other.

Cor. Two parallels, AB, CD, comprehended between two other parallels, AC, BD, are equal; and the diagonal BC di vides the parallelogram into two equal triangles.

PROPOSITION XXX.

THEOREM (Converse of Prop. XXIX.). If the opposite sides of a quadrilateral are equal, each to each, the equal sides are parallel, and the figure is a parallelo gram.

Let ABDC be a quadrilateral, having its A

B opposite sides equal to each other, viz. : the side AB equal to CD, and AC to BD; then will the equal sides be parallel, and the figure will be a parallelogram.

D Draw the diagonal BC; then the triangles ABC, BCD have all the sides of the one equal to the corresponding sides of the other, each to each; therefore the angle ABC is equal to the angle BCD (Prop. XV.), and, consequently, the side AB is parallel to CD (Prop. XXII.). For a like reason, AC is parallel to BD; hence the quadrilateral ABDC is a parallelogram. Therefore, if the opposite sides, &c.

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If two opposite sides of a quadrilateral are equal and parallel, the other two sides are equal and parallel, and the figure is a parallelogram.

B

С

Let ABDC be a quadrilateral, having the A sides AB, CD equal and parallel ; then will the sides AC, BD be also equal and parallel, and the figure will be a parallelogram.

D Draw the diagonal BC; then, because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal (Prop. XXIII). Also, because AB is equal to CD, and BC is common to the two triangles ABC, BCD, the two triangles ABC, BCD have two sides and the included angle of the one, equal to two sides and the included angle of the other; therefore, the side AC is equal to BD (Prop. VI.), and the angle ACB to the angle CBD. And, because the straight line BC meets the two straight lines AC, BD, making the alternate angles BCA, CBD equal to each other, AC is parallel to BD (Prop. XXII.); hence the figure ABDC is a parallelogram. Therefore, if two opposite sides, &c.

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D

The diagonals of every parallelogram bisect each other
Let ABDC be a parallelogram whose di-

A

B agonals, AD, BC, intersect each other in E; then will AE be equal to ED, and BE to EC.

Because the alternate angles ABE, ECD C are equal (Prop. XXIII.), and also the alternate angles EAB, EDC, the triangles ABE, DCE have two angles in the one equal to two angles in the other, each to each, and the included sides AB, CD are also equal; hence the remaining sides are equal, viz.: AE to ED, and CE to EB. Therefore, the diagonals of every parallelogram, &c.

Čor. If the side AB is equal to AC, the triangles AEB, AEC have all the sides of the one equal to the corresponding sides of the other, and are consequently equal ; hence the angle AEB will equal the angle AEC, and therefore the di agonals of a rhombus bisect each other at right angles.

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