235. If a point be taken on the radius of a circle, and this radius be then produced, and a second point be taken on it, without the circumference of the circle, these points being so situated, that the radius of the circle shall be a mean proportional between their distances from the centre, then, if lines be drawn from these points to any point of the circumference, the ratio (of these lines) will be constant.

:

Let P be the point within the circumference, and Q the point without; then if CP: CA: CA: CQ, the ratio of QM and MP will be the same, for all positions of the point M.

For by hypothesis, CP: CA :: CA: CQ; or substituting CM for CA, CP: CM :: CM : CQ; hence the triangles CPM, CQM, have each an equal angle Q C contained by proportional sides; hence they are similar (208.); and hence the third side

M

A P

M

MP is to the third side MQ, as CP is to CM or CA. But by division, the proportion CP: CA:: CA: CQ gives CP: CA:: CA-CP : CQ-CA, or CP : CA :: AP : AQ; therefore MP MQ :: AP: AQ.

13

PROBLEMS RELATING TO THE THIRD BOOK.

PROBLEM.

236.

To divide a given straight line into any number of equal parts, or into parts proportional to given lines.

First. Let it be proposed to divide the line AB into five equal parts. Through the extremity A, draw the indefinite straight line AG; and taking AC of any magnitude, apply it five times upon AG; join the last point of division G and the extremity B, by the straight line GB; then draw CI parallel to F GB: AI will be the fifth part of the line AB; G and thus, by applying AI five times upon AB, the line AB will be divided into five equal parts.

For since CI is parallel to GB, are cut proportionally in C and I. of AG, hence AI is the fifth part of Secondly. Let it be proposed to divide the line AB into parts proportional to the given lines P, Q, R. P Through A, draw the indeQ finite line AG ; make AC= P, CD=Q, DE=R; join the extremities E and B;

R

E

-I

M

B

the sides AG, AB (196.) But AC is the fifth part

AB.

A

and through the points C, D draw CI, DF parallel to EB; the line AB will be divided into parts AI, IF, FB proportional to the given lines P, Q, R.

For, by reason of the parallels CI, DF, EB, the parts AI, IF, FB are proportional to the parts AC, CD, DE; and by construction, these are equal to the given lines P, Q, R.

237. To find a fourth proportional to three given lines A, B, C.

parallel to AC; DX will be the fourth proportional required: for since BX is parallel tó AC, we have the proportion DA: DB:: DC: DX; now the three first terms of this proportion are equal to the three given lines; consequently DX is the fourth proportional required.

238. Cor. A third proportional to two given lines A, B, be found in the same manner, for it will be the same as a fourth proportional to the three lines A, B,

may

B.

239. To find a mean proportional between two given lines A and B.

Upon the indefinite line DF, take DE A, and EF-B; upon the whole line DF, as a diameter, describe the semicircle DGF; at the D point E, erect upon the diameter the perpendicular EG meeting the cir- Bcumference in G; EG will be the A mean proportional required.

For the perpendicular EG, let fall from a point in the circumference upon the diameter, is a mean proportional between DE, EF, the two segments of the diameter (215.); and these segments are equal to the given lines A and B.

240. To divide a given line into two parts, such that the greater part shall be a mean proportional between the whole line and the other part.

Let AB be the given line.

At the extremity B of the line AB, erect the perpendicular BC equal to the half of AB; from the point C as a centre, with the radius CB describe a semicircle;

draw AC cutting the circumfer

ence in D; and take AF=AD: A

the line AB will be divided at the point F in the manner required; that is, we shall have AB: AF:: AF : FB.

For, AB being perpendicular to the radius at its extremity, is a tangent; and if AC be produced till it again meets the circumference in E, we shall have (228.) AE :AB::AB: AD; hence, by division, AE-AB: AB:: AB-AD : AD. But since the radius is the half of AB, the diameter DE is equal to AB, and consequently AE-AB=AD=AF; also, because AF AD, we have AB—AD=FB; hence AF: AB :: FB: AD or AF; whence, by inversion, AB: AF :: AF: FB.

241. Scholium. This sort of division of the line AB is called division in extreme and mean ratio: the use of it will be perceived in a future part of the work. It may further be observed, that the secant AE is divided in extreme and mean ratio at the point D; for, since AB=DE, we have AE: DE :: DE: AD.

PROBLEM.

242. Through a given point, in a given angle, to draw a line so that the segments comprehended between the point and the two sides of the angle, shall be equal.

Let BCD be the angle, and A the point.

Through the point A draw AE parallel to CD, make BE-CE, and through the points B and A draw BAD; this will be the line required.

For, AE being parallel to CD, we have BE EC: BA: AD; but BE= EC; therefore BA=AD.

B

E

PROBLEM.

D

243. To describe a square that shall be equivalent to a given parallelogram, or to a given triangle.

First. Let ABCD be the given parallelogram, AB its base, DE its altitude between AB and DE find a mean proportional XY; then will the square constructed upon

XY be equivalent to the parallelogram ABCD.

For by construction, AB : XY :: XY: DE; therefore XY2=AB.DE; but AB.DE is the measure of the parallelogram, and XY that of the square; consequently they are equivalent.

Secondly. Let ABC be the given triangle, BC its base, AD its altitude: find a mean proportional between BC and the half of AD, and let XY be that mean; the square constructed upon XY will be equivalent to the triangle ABC.

For since BC: XY:: XY: AD, it follows that XY2== BC. AD; hence the square constructed upon XY is equivalent to the triangle ABC.

PROBLEM.

244. Upon a given line, to describe a rectangle that shall be equivalent to a given rectangle.

Let AD be the line, and ABFC the given rectangle.