Construction. By the pre

vious construction make a rectangle equal to ABCDE, and let PQRS be the rectangle so made.

Then if PQ = QR the rectangle is a square; but if

P

R

not, produce PQ to T, making QT= QR; on PQT as diameter describe a semicircle, U being the centre, and produce RQ to meet the circumference in V.

If a square be described on QV, this square will be equal to ABCDE.

Proof. For since PQ is the sum of PU and UQ, and QT is the difference of PU (or UT) and UQ, it follows (from II. 8) that the rectangle PQ × QT= PU2 – UQ3; but PU2 = UV3, and therefore PUa – UQ2 = UV2 – UQ3, that is, VQ, by 11. 9, Cor.

But the rectangle PQ× QT is the rectangle PQRS, which was made equal to ABCDE.

=

Therefore VQ ABCDE, and the square described on VQ is the square required*.

Remark. If the given figure is not rectilineal, it cannot be divided into triangles: hence it is impossible by this method to construct a square equal to a given curvilinear area. Nor can any method depending on the use of the ruler and compasses only (see p. 60), construct a square equal to some curvilinear areas, such as the circle. This is the problem of squaring the circle, the solution of which cannet be effected without the use of other instruments.

* Euclid, II. 14.

PROBLEM V.

To construct a rectilineal figure equal to a given rectilineal figure and having the number of its sides one less than that of the given figure; and thence to construct a triangle equal to a given rectilineal figure.

Let ABCDE be the given rectilineal figure. Join AC, and through B draw BK parallel to AC to meet DC produced in K. Join AK.

Then since the triangle ABC is equal to the triangle AKC, being on the same base AC and between the same parallels; add to each ACDE; therefore the figure ABCDE is equal to the figure AKDE, which has the number of its sides diminished by one.

Since this process can be repeated any number of times it is evident that any polygon can be reduced in this manner to an equivalent triangle.

PROBLEM VI.

To divide a straight line, either internally or externally, into two segments such that the rectangle contained by the given line and one of the segments may be equal to the square on the other segment.

Let AB be the given line.

First, to divide it internally.

F

G

Construction. Draw a square ACDB on AB; bisect AC in E. Join BE; produce EA to F, making EF= EB; on AF A describe a square AFGH.

AH and HB are the parts required, so that the rectangle AB × BH=AH3,

Proof. Produce GH to meet CD in K.

E

B

Then since CA is bisected in E, and divided externally in F,

therefore CF × FA = EF2 - EA2 (11. 8. Cor.);

but EF = EB, and therefore EFo – EA2 – ABa
(II. 9. Cor.),

therefore CFX FA=AB2;

that is, the figure FK = the figure AD; take from each AK, and therefore

FH= HD.

But HD is the rectangle AB× BH; and FH is the square on AH;

therefore AB × BH = AH2.*

COR. The line FA is divided externally in C, so that
FC. FA=CA2.

* Euclid, II. II.

Secondly, to divide it externally.

K H

F

Construction.

Produce BA, and take AH equal to AB. Bisect AH in E. On AB describe a square ACDB. Join EC; from EH produced, cut off EK= EC.

Then will AB. BK=AK

Proof. On KA describe a square KGLA, and produce DC to meet KG in F.

but

Then, since AH is bisected in E and produced to K,

therefore

AKX KH=EK' - EA';

EK-EC, and therefore EK — EA2 = AC2,

AKx KH-AC;

(II. 8.)

that is, the square on AK is equal to the rectangle AB×BK.

EXERCISES.

I.

Construct a square double of a given square.

Q. E. D.

2.

Construct a square equal to two, or three, or any number of given squares.

3.

Divide a straight line into two parts, such that the square of one of the parts may be half the square on the whole line.

4. Given the base, area, and one of the angles at the base, construct the triangle.

5. Find the locus of a point which moves so that the sum of the squares of its distances from two given points is constant.

We subjoin a few problems and theorems as miscellaneous exercises in the Geometry of angles, lines, triangles, parallelograms, and the equality of areas.

MISCELLANEOUS THEOREMS AND PROBLEMS.

I. Prove that the acute angle between the bisectors of the angles at the base of an isosceles triangle is equal to one of the angles at the base of the triangle.

2. Find a point equally distant from three given straight lines.

3. If the diagonals of a quadrilateral bisect one another and are equal to one another, the figure will be a rectangle.

4. If the diagonals of a quadrilateral bisect one another at right angles and are also equal, the figure will be a square.