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PROP. XIII. THEOR.
If from a point without a straight line a perpendicular be drawn to the
line, and different straight lines making oblique angles with the straight line, the perpendicular will be shorter than either of the oblique lines. Two obliques equally distant from the perpendicular will be equal, and an oblique farther from the perpendicular, will be longer than one nearer thereto.
Let A be a point without the straight line DE, AB perpendicular to DE, AC and AE obliques, equidistant from AB, and AD farther than AC from AB. AB will be less than AE, AC, or AD; AC and AE will be equal, and AD greater than AC. Produce AB to F, make BF equal to AB, and join AC, AD, AE, FC, and FD. Because ABC is a right angle, FBC is also a right angle (Cor. 3. 1.), and the tri
F angles ABC, FBC have the side AB equal to the side BF, the side CB common, and the contained angle ABC equal to the contained angle FBC; therefore the triangles are equivalent (4. 1.), and the side AC is equal to the side FC. But AF is a straight line, and therefore it is less than ACF (Def. 16.); and therefore AB, the half of AF, is less than AC, the half of ACF; therefore the perpendicular is shorter than any oblique.
Because the triangles ABC, ABE have the side BC equal to the side BE, the side AB common, and the contained angle A BC equal to the contained angle ABE, being both right angles by the hypothesis, (Cor. 2. 1.), the triangles are equivalent, and the side AC is equal to the side AE ; therefore two obliques equidistant from the perpendicular, are equal.
In the same way it may be shewn, that the triangle ABD is equivalent to the triangle FBD, and the side AD equal to the side FD. Then the straight lines AC, FC are drawn to the point C, within the triangle ADF; therefore, they are less than AD and FD (13. 1.), and therefore AC, the half of AC, FC, is less than AD, the half of AD, FD; and therefore the oblique farther from the perpendicular, will be greater than one nearer thereto. Therefore, If, &c. Q. E. D.
Cor. A perpendicular from a point to a straight line, measures the shortest distance between the point and the line.
PROP. XIV. PROB.
At a given point in a given straight line, to make an angle equal to e
given angle. Let B be the given point in the given straight line AB, and let DEF be the given angle; it is required to make at B an angle with the line
AB, equal to DEF. From E, with any radius ED, describe the arc DF, and from B, with the same radius, describe the arc AC; join DF, and from A, with the radius DF, describe the arc CG, intersecting the arc AC at C, and join BC; the angle ABC is equal to the angle DEF.
Join AC; because the triangles ABC, DEF, have two sides AB, BC equal to the two sides DE, EF, and the third side AC equal to the third side DF; therefore the angle ABC is equal (7. 1.) to the angle DEF. Therefore, at the given point B, in the given straight line AB, the angle ABC is made equal to the given angle DEF. Which was to be done.
Two straight lines in the same direction from different points, may
be called parallel lines.
II. A plane figure enclosed by four straight lines, may be called a quadrangle.
III. A quadrangle, whose opposite sides are parallel, may be called a
A parallelogram, whose angles are right angles, may be called a
be called a rhombus.
IX. A straight line joining the two opposite angles of a parallelogram, may be called a diagonal.
D a diagonal. Through K, any point in AC, draw EF parallel to AD; and also GH parallel to AB. A А
F parallelogram may be named by the letters at two opposite angles, as AC.
the diagonal passes, may be said
XII. The remaining parallelograms, as EG, HF, may be called the complements.
XIII. The complements, and one of the parallelograms about the diagonal, may be said to make a gnomon.
XIV. A gnomon may be expressed by the letters at the opposite angles, as AGF.
XV. A rectangle may be said to be contained by the two lines which contain one of the angles.
XVI. A perpendicular drawn from an angle of a triangle or parallelogram,
to the base or base produced, may be called the altitude of the triangle, or parallelogram.
PROPOSITION I. THEOREM.
If a straight line falling on two other straight lines, makes an angle with one of the lines equal to an angle which it makes with the other, the remaining angles which it makes with the first line, are equal to the remaining angles which it makes with the other line, each to each.
Let the straight line AB, falling on the straight lines CD, EF, make the angle AGD equal to the angle GHF; the angle AGC will be equal to the angle GHE, the angle EHB to CGH, and BHF to DGH.
Because AG and CD are straight lines, the angles AGD and AGC are together equal (3 cor. 1. 1.) to two right angles. For the same reason, the angles GHF and GHE are together equal to two right angles ; take away the equal angles AGD and GHF, and the remainder AGC is equal to the remainder GHE: and because AGC and DGH are vertical angles, they are equal (3. 1.); but AGC has been proved to be equal to GHE, and therefore DGH is equal to GHE; and because GHE and FHB are vertical angles, they are equal (3.1.); and therefore DGH is equal to BHF: and in the same way it may be provс C
ed that EHB is equal to CGH. Therefore, if a straight line falling on two other straight lines, makes an angle with one of the lines equal to an angle which it makes with the other, the remaining angles which it makes with the first line, are equal to the remaining angles which it makes with the other line, each to each. Which was to be proved.
Cor. The similarly placed angles AGD and GHF, and their verticals CGH and BHE, may be called corresponding angles. Likewise AGC and GHE, and their verticals DGH and BHF.
PROP. II. THEOR.
If a straight line, falling on two other straight lines in the same plane,
makes the corresponding angles equal, any other straight line falling on those two straight lines, will also make the corresponding ungles equal.
Let the straight line EF fall on the straight lines AB and CD, and make the corresponding angles equal; any other straight line, as GH, will also make the corresponding angles equal.
Because the angles which EF makes with AB, are equal to the angles which it makes with CD, the difference in direction between