« AnteriorContinuar »
Let AB be a given straight line.
as centres, with 3 radius greater than one half of AB, describe arcs intersecting at E and F: join E and F, by the straight fine EF. Then · will EF bisect the given line AB. For, E and F are each equally distant from A and B; and consequently, the line EF bisects AB (B. I., P. XVI., C.).
To erect a perpendicular to a given straight line, at a given
point of that line.
Let EF be a given line, and let 4 be a given point o that line.
From A, lay off the equal distances AB and 'AC; from B and C, as centres, with a radius greater than one half E
of BC, describe aics intersecting at D; draw tłe line AD: then will AD be the perpendicular required. For, D and A are each equally distant from B and C; consequently, DA is perpendicular to BC at the given point A (B. I., P. XVI., C.).
To draw a perpendicular to a given straight line, from a
given point without that line.
Let BD be the given line, and the given point. From A, as a centre, with a ra
A dius sufficiently great, describe an arc cutting BD in two points, Band D; with Band D as centres, and a radius greater than one-half of BD, describe arcs intersecting at E; draw AE: then will AE be the perpendicular required. For, A and E are each equally distant from B and hence, AE is perpendicular to BD (B. I., P. XVI., C.).
At a point on a given straight line, to construct an angle
equal to a given angle.
Let A be the given point, AB the given line, and IKL the given angle. From the vertex K
D centre, with any radius KI, describe the arc IL, terminat
B ing in the sides of the angle. From A as a centre, with a radius AB, equal to KI,
describe the indefinite arc BO; then, with a radius equal to the chord LI, from B a centre, describe an arc cutting the arc BO in D; draw AD: then will BAD be equal to the angle K. For, the BD, IL K
B bave equal radii and equal chords : hence, they are equal (P. IV.); therefore, the angles BAD, IKL, measured by them, are also equal (P. XV.).
To bisect a given arc, or a given angle.
10. Let AEB be a given arc, and Cits centre,
Draw the chord AB; through C, draw CD perpendicular to AB (Prob. III.) : then will CD
bisect the arc AEB (P. VI.).
2°. Let ACB be a given angle.
B With c as a centre, and any radius CB, describe the arc BA ; bisect it by the line CD, as just explained : then will CD bisect the angle ACB. .
For, the arcs AE and EB are equal, from what was jnst shown ; consequently, the angles ACE and ECB are also equal (P. XV.).
Scholium. If each half of an arc or angle be bisected, the original arc or angle will be divided into four equal parts; and if each of these be bisected, the original arc or angle will be divided into eight equal parts ; and so on.
Through a given point, to draw a straight line parallel to
a given straight line.
Let A be a given point, and BC a given line.
From the point A as a centre, with a radius AE, greater than the F
B shortest distance from A to BC, describe an indefinite arc EO; from E as a centre, with the same dius, describe the arc AF;- lay off ED equal to AF, and draw AD: then will AD be the parallel required.
For, drawing AE, the angles AEF, EAD, are equal (P. XV.); therefore, the lines AD, EF are parallel (B. I., P. XIX., C. 1.).
Given, two angles of a triangle, to construct the third
Let A and B be given angles of a triangle.
Draw a line DF, and at some point of it, as E, construct the angle FEII equal to A, and HEC equal to B. Then, will CED be
D equal to the required angle.
For, the sum of the three angles at E is equal to two right angles (B. I., P. I., C. 3), as is also the sum of the three angles of a triangle (B. I., P. XXV.). Consequently, the third ang!e CED must be equal to the third angle of the triangle.
Given, two sides and the included angle of a triangle, 10
construct the triangle.
Let B and C denote the given sides, and a the given angle.
Draw the indefinite line DF, and at D construct an angle
B side C, and on DE, lay off
C DG equal to the side B; draw GH: then will DGH be the required triangle (B. I., P. V.).
Given, one side and twon angles of a triangle, to construct
The two angles may be either both adjacent to the given side, or one may be adjacent and the other opposite to it. In the latter case, construct the third angle by Problem VIL We shall then have two angles and their included side.
Draw a straight line, and on it lay off DE equal to the given side ; at D construct an angle equal to one of the adjacent angles, and at E construct an angle equal to the other adjacent angle; produce the sides DF and EG till they intersect at I : then will DEH be the triangle required (B. I, P. VI.).