Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

the angle CDF from four right-angles; that is, we must consider this angle as exceeding two right-angles.

For, drawing CD' and FD' respectively parallel to DF and DC, we shall form a parallelogram CDFD'; and the figure ABCD/FG will be a convex polygon, having the same number of sides as the original polygon. And it is moreover obvious that the sum of all the interior angles of this convex

A

D'

F

B

polygon is the same as the sum of all the angles of the original polygon, if we consider the angle at D as what remains after taking the angle CDF from four right angles.

(42.) A practical application of the principle contained in this proposition is made by the land surveyor, as a test of the accuracy of his work. By means of the courses of the different sides of his field, he can determine the magnitude of all the interior angles, estimated in degrees and minutes: he takes the sum of all these angles, and observes whether it is such as to correspond with the above conditions. If the number of sides of the field be denoted by n, then the sum of all the interior angles ought to be equal to (2 n-4) times a right-angle; thus, if there are 7 sides in the field, all the angles ought to amount to 10 right-angles, or 9000. If the field contain re-entering angles (40,) it will be necessary to modify somewhat this method.

PROPOSITION XXV.

THEOREM. When one side of a triangle is produced, the exterior angle is equal to both the interior and opposite angles taken together.

Let the side AB of the triangle ABC be produced to D; then will the exterior angle CBD be equal to the sum of the two interior and opposite A angles A and C.

B

D

For, conceive BF to be drawn parallel to AC; then BC, meeting the two parallel lines AC, BF, makes the alternate angles C and CBF equal, (Prop. xvII;) and AD, cutting the same two parallels AC, BF, makes the interior and exterior angles on the same side, A and FBD, equal to each other, (Prop. xx ;) therefore, by equal additions, the sum of the two angles A and C is equal to the sum of the angles CBF and FBD, that is, to the whole angle CBD, (Ax. II.)

PROPOSITION XXVI.

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

If AB, AC, AD, be lines drawn from the given point A, to the indefinite line DF, of which AB is perpendicular; then will the perpendicular AB be less than AC, and AC less than AD.

D

A

B

For, the angle B being right, the angle C must be acute, (Prop. vI,) and therefore less than the angle B; but the less side of a triangle is situated opposite the less angle, (Prop. xv,) 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, (Prop. 1 ;) consequently the angle ADC is acute, (Prop. vi,) and therefore it is

less than the angle ACD; and since the less side is opposite the lesser angle, the side AC must be less than the side AD.

Cor. A perpendicular measures the shortest distance of a point from a line.

PROPOSITION XXVII.

THEOREM. The opposite angles and sides of a 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 AC; then will its opposite sides and angles be equal to each other, and the diagonal will divide it into two equal triangles.

A

D

B

angle DCA, hence the two

equal to two

For, since the sides AB and DC are parallel, as also the sides AD and BC, (Def. XVII,) and the line AC meets them, the alternate angles are equal, (Prop. XVII,) namely, the angle BAC to the and the angle BCA to the angle DAC; triangles, having two angles of the one angles of the other, have also their third angles equal, (Prop. xxiv, Cor. 1,) namely, the angle B equal to the angle D, which are two of the opposite angles of the parallelogram. Also, if to the equal angles BAC, DCA be added the equal angles DAC, BCA, the wholes will be equal, (Ax. II,) namely, the whole angle BAD to the whole angle DCB, which are the other two opposite angles of the parallelogram.

Again, since the side AC is common to the two triangles ACB and CAD, we have two angles and the interjacent side of the one equal to two angles and the interjacent side of the other, and therefore the triangles are equal, (Prop. IV.) Therefore the side AB is equal to its opposite side CD, and BC equal to its opposite side DA, and the whole triangle ABC equal to the whole triangle BCD.

Cor. 1. If one angle of a parallelogram be a rightangle, all the other three angles will also be right-angles, and the parallelogram will be a rectangle, (Def. XVIII.)

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

PROPOSITION XXVIII.

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

Let ABCD be a quadrilateral, having the opposite sides equal, namely, the side AB equal to DC, and AD equal to BC; then will these equal sides be also

A

D

с

parallel, and the figure will be a parallelogram.

B

For, let the diagonal AC be drawn; then the triangles ABC, CDA, being mutually equilateral, are also mutually equiangular, (Prop. vIII,) or have their corresponding angles equal; consequently the opposite sides are parallel, (Prop. xvIII,) namely, the side AB parallel to DC, and AD parallel to BC, and the figure is a parallelogram, (Def. XVII.)

PROPOSITION XXIX.

THEOREM. 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 corresponding extremes, be also equal and parallel.

A

D

B

For, draw the diagonal AC; then, because AB and DC are parallel, the angle BAC is equal to its alternate angle DCA, (Prop. xvII;) hence the two triangles, having two sides and the included angle of the one equal to two sides and the included angle of the other, namely, the side equal to the side CD, the side AC common, and the contained angle BAC equal to the contained angle DCA, have also the remaining sides and angles respectively equal, (Prop. III;) consequently AD is equal to BC, and also parallel to it, (Prop. xvIII.)

[blocks in formation]

it is obvious that AB must in every position be parallel to DC, since AB and DC join the corresponding extremes of the two equal and parallel lines.

« ΠροηγούμενηΣυνέχεια »