Let X Y be the given straight line and A the given point.

Take the points B and C on XY at equal distance from the point A; from the centres B and C, with a radius greater than half of B C,

describe two circumferences: they will cut each other in a point D [62]. Draw a straight line through the points D and A: the line A D is the required perpendicular.

Because the point D belongs to both the equal circumferences B and C, its distance from the centres B and C is equal to the common radius; therefore the point D is equally distant from the points B and C. Because AB and AC are equal to each other [Const.], the line AD has the points A and D, equally distant from the points B and C; thence A D is the middle perpendicular to A B [14]: that is, A D is a perpendicular to XY: therefore through the point A given in the given straight line X Y, a perpendicular has been drawn, W. W. T. B. D.

PROBLEM 6. To draw the middle perpendicular to a given straight line.

PROBLEM 7. To bisect a given straight line.

PROBLEM 8. To bisect a given arc.

PROBLEM 9.

Through a point given without a given straight line, to draw a perpendicular to that line.

Let XY be the given straight line and A the given point.

X

B

D

A

C

From the centre A, with a radius greater than the distance of A to XY, describe a circumference cutting X Y in two points B and C; from the centres B and C, with a radius greater than half of B C describe two circumferences cutting each other in D, and through the points A and D draw a straight line: the line AD is the required perpendicular.

Because B and C are points of the same circumference, their distances from the centre A are equal to each other; therefore the point A is equally distant from B and C. Because the point D belongs to both the equal circumferences B and C, its distances from the centres B and C are equal to each other; thence the line A D bas two points A and D equally distant from the points B and C, and

is the middle perpendicular to B C [14]: therefore AD is perpendicular to X Y,' which W. T. B. D.

PROBLEM 10. Through a point given without a given straight line, to draw a parallel to that line.

PROBLEM 11.

To divide a given straight line into a given number of equal parts.

Let A B be the given straight line to be divided into five equal parts.

Through the point A, draw the straight line AC in any direction; take any distance A D and set it off upon A C five times successively in AD, DE, EF, FG and GH; join HB, and through the points G, F, E and D draw G I, F K, E L and D M parallel to HB [Probl. 10]: the line A B is divided by the points M, L, K and I into five equal parts.

Let AN be a parallel to H B. Because the intercepts of A C between parallels HB, GI, FK, EL, D M, and AN, are equal to each other [Const.], they are equally distant from each other [24 ii]; thence the intercepts of AB between the same parallels are equal to each other [231]: therefore the straight line AB

1 As in the theorems, the student must supply here, and in the demonstration of most other problems, the rest of the reasoning suppressed for the sake of brevity. Also the student should find, for each problem, other solutions than those given in The Elements.

is divided by the points M, L, K and I into five equal parts, W. W. T. B. D.

PROBLEM 12.

(Discussed.)

Through a given point, to draw a transversal to two given parallels between which the intercept shall be equal to a given straight line.

Let O be the given point, A B and C D the given parallels, and EF

the straight line to which the intercept of the required transversal

must be equal.

From any point H of A B, with a radius equal to E F, describe a circumference cutting CD in a point I; join HI, and through the point O draw OK parallel to HI [Probl. 10]: the line OK is the required transversal.

Let LK be the intercept of OK between the parallels A B and CD. Because OK and HI are parallel to each other [Const.], the intercepts L K and HI between the parallels A B and C D are equal to each other [22]. But HI is equal to EF [Const.]; then LK is equal to EF [i]: therefore OK is a transversal, the intercept LK of which, between the parallels A B and CD, is equal to EF, which W. T. B. D.

The given straight line E F may be equal to the distance of the parallels AB and CD from each other, or it may be greater, or smaller.

Let HG and OM be perpendicular to CD: then, they are also perpendicular to AB [17]. If E F is equal to the distance of the parallels A B and C D from each other, it is equal to H G [11,] then the circumference described from the point H with a radius equal to EF will be tangent to CD in the point G [52]: but the intercept of OM between A B and C D is equal to H G [19]; therefore O M will be the only transversal satisfying the data.

If E F is greater than the distance of A B and C D from each other it is greater than H G [11], and the circumference described from the point H, with a radius equal to EF, will cut CD in two points I and J; therefore, if HI and H J be joined, and parallels OK and ON to them be drawn through the point O, there will be two transversals, OK and O N, satisfying the data of the problem.

If E F is smaller than the distance of AB and CD from each other, it is smaller than HO [11], then the circumference described from the point H with a radius equal to E F will not touch the line CD: therefore there will be no transversal satisfying the data of the problem; that is, the problem will be impossible.

Therefore the problem may have two solutions, or one only, or

none.