GEOMETRICAL PROBLEMS.

PROBLEM I.

To make a triangle of three given right lines BO, LB, LO, of which any two must be greater than the third. fig. 42.

Lay BL from B to L; from B with the line BO, describe an arc, and from L with LO describe another arc; from 0, the interjection point of those arcs, draw BO and OL, and BOL is the triangle required.

This is manifest from the construction.

PROB. II.

At a point B in a given right line BC, to make an angle equal to a given angle A. fig. 43.

Draw any right line ED to form a triangle, as EAD, take BF AD, and upon BF make the triangle, BFG, whose side BG=AE, and GF=ED [by the last] then also the angle B=A if we suppose one triangle be laid on the other, the sides will mutually agree with each other, and therefore be equal; for if we consider these two triangles are made of the same given three lines, they are manifestly one and the same triangle.

Otherwise,

Upon the centres A and B, at any distance, let two arcs, DE, FG, be described; make the arc FG=DE, and thro' B and G draw the line BG, and it is done.

For since the chords ED, GF, are equal, the angles A and B are also equal, as before [by def. 19.3

PROB. III.

To bisect or divide into two equal parts, any given right lined angle, BAC. fig. 44.

In the lines AB and AC, from the point A set off equal distances AE=AD, then, with any distance

each other in some point F; and the right line AF, joining the points A and F, will bisect the given angle BAC.

For if DF and FE be drawn, the triangles ADF, AEF, are equilaterial to each other, viz. AD=AE, DF, FE, and AF, common, wherefore DAF-EAF, as before.

With any distance, more than half the line, front A and B, describe two circles CFD, CGD, cutting each other in the points C and D ; draw CD, intersecting AB in E, then, AE=EB.

For, if AC, AD, BC, BD, be drawn the triangles ACD, BCD, will be mutually equilateral and consequently the angle ACE-BCE: therefore the triangle ACE, BCE, having AC-BC, CE, common, and the angle ACE=BCE; [by theo. 6.] the base AE the base BE.

Cor. Hence it is manifest, that CD not only bisects AB, but is perpendicular to it. [by def. 11.]

PROB. V.

On a given point A, in a right line EF, to erect a perpendicular. fig. 46.

fig. B

E

From the point A lay off on each side, the equal distances, AC, CD; and from C and D, as centres, with any interval greater than AC or AD, describe two arcs intersecting each other in B; from A to B draw the line AB, and it will be the perpendicular required.

For let CB, and BD be drawn; then the triangles CAB, DAB, will be mutually equilateral and equiangular, so CAB=DAB, a right angle, [by def. 11.]

PROB. VI.

To raise a perpendicular on the end B of a right line AB. fig. 47.

From any point D not in the line AB, with the distance from D to B, let a circle be described cutting AB in E; draw E thro' D the right line EDC, cutting the periphery in C, and join CB; and that is the perpendicular required.

EBC being a semicircle, the angle EBC will be a right angle [by cor. 5. theo. 7.]

PROB. VII.

From a given point A, to let fall a perpendicular upon a given right line BC. fig. 48.

From any point D, in the given line, take the distance to the given point A, and with it describe a circle AGE, make GE=AG, join the points A and E, by the line AFE, and AF will be the perpendicular required.

Let DA, DE, be drawn; the angles ADF=FDE, DA DE being radii of the same circle, and DF, common; therefore [by theo. 6.] the angle DFA =DFE, and FA a perpendicular. [By def. 11.]