Sidebilder
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shortest distance to the line, draw an arc cutting AB in D and E. Bisect DE in F (Prob. I). Then draw CF, and it will be the perpendicular required.

Join CD and CE. Now, since the triangles DCF, ECF, are mutually equilateral, the angles CFD, CFE, are equal (Theo. XII); they are, therefore, right angles, and CF is a perpendicular to AB from the point C.

PROBLEM IV.

To bisect a given angle.

Solution. Let BAC be the given angle. From A as a center, with any radius, draw an arc BDC intersecting both sides of the angle. Then draw B AD perpendicular to the chord BC (Prob. III), and it will bisect the angle.

A

C

For, since the perpendicular AD is radius, it bisects the arc BDC (Theo. XXV). Therefore, the arcs BD, DC, being equal, the subtended angles BAD, DAC, must also be equal (Theo. XXIV), and the angle BAC is bisected by AD.

PROBLEM V.

At a given point in a straight line, to make an angle equal to a given angle.

Solution. Let ACB be the given angle, and F the given point in a straight line FE. From

C

BF

C, as a center, with any radius as CB, describe an are BA intersecting both the sides of the angle; also, draw the chord BA. Then, from F as a center, with

a radius FE equal to CB, draw an indefinite arc; and from the center E, with a radius equal to the chord BA, draw an arc intersecting the other at D; also, join DE and DF. The angle DFE is the angle required. For, by equality of the triangles DEF, ABC, the angle F is equal to the angle C (Theo. XII).

PROBLEM VI.

Through a given point to draw a straight line parallel to a given line.

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CD. Through C draw EF, making the angle ECD equal to the angle CDB (Prob. V). These being alternate angles, the straight line EF must be parallel to AB (Cor. 1, Theo. III).

PROBLEM VII.

Two angles of a triangle being given, to find the third.

Solution. At any point C in

a straight line AB, make ACD equal to one of the given angles (Prob. V), and DCE equal to the other; then will

D

E

B

ECB be equal to the third angle. For the sum of the three angles at C is two right angles (Cor. 1, Theo. I); which is also the sum of the three angles of a triangle (Theo. V).

PROBLEM VIII.

Given the hypotenuse and one side of a right-angled triangle, to construct the triangle.

Solution. Draw AB equal to the given side. At B erect a perpendicular BC (Prob. II). From A as a center, with a radius equal to the given hypotenuse, describe an arc intersecting BC at D. Join AD.

B

It is evident that ABD is the triangle required.

PROBLEM IX.

To draw a circle through three given points.

Solution. Let A, B, and C be the given points. Draw the straight lines AB and BC, and from their middle points erect the perpendiculars DF and EF. Also, join AF, BF, and CF.

D

B

Now, because the triangles AFD, BFD, have the side AD equal to the side BD, and the side DF common, and the included angle ADF equal to the included angle BDF, the third side AF is equal to the third side BF (Theo. VII). In the same manner it may be shown that BF is equal to CF. Therefore, the three lines AF, BF, and CF are equal; and if from F as a center, with one of these equals as radius, a circle be drawn, it will pass through the three points A, B, and C.

Cor. Hence, the center of a given circle may be found by erecting perpendiculars on the middle points. of two chords, and producing them till they meet.

EXERCISES.

1. Given the three sides of a triangle, to construct the triangle.

2. Given two angles and the included side, to construct the triangle.

3. Given two adjacent sides and the included angle of a parallelogram, to construct the parallelogram. 4. To draw a diameter of a given circle.

5. At a given point in the circumference of a circle, to draw a tangent to the circle.

PLANE GEOMETRY.

BOOK II.

PROPORTIONS OF MAGNITUDES LYING IN THE SAME PLANE.

SEC. X.-OF POLYGONS.

DEFINITIONS.

1. In mutually equiangular figures, the sides which are similarly situated with respect to the equal angles, called homologous

are

sides.

Thus,

If the angles A, B, C, are respectively equal to the angles a, b, c, the side AB is homol

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ogous to the side ab, the side BC to the side be, and CA to ca.

Two triangles or other polygons are called similar when they are mutually equiangular, and their homologous sides are proportional.

2. Magnitudes are called proportionals when the first has the same ratio to the second, that the third has to the fourth, the fifth to the sixth, etc.

The first terms of the several equal ratios are called the antecedents, and the second the consequents. Thus, If A B C D E F, the terms A, C, E, are the antecedents, and B, D, F, the consequents.

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