But figure FK is rect. DF, FA, since FG= FA.

From these equals take away the fig. AK which is common; then

fig. FH fig. HC.

But fig. FH is a square, and s.. the sq. on AH, and fig. HC is equal to rect. AB, HB, for BC = AB.

Hence

sq. on AH=rect. AB, BH.

Def. When a straight line is divided into two parts so that the rectangle contained by the whole line and one of the parts is equal to the square on the other part, the line is said to be divided in 'medial section.' The line is also said to be divided in 'extreme and mean ratio,' for in this case, as will be seen in Book VI, the ratio of the whole line to one part is equal to the ratio of that part to the other.

The analysis [see page 101] of this problem will show how the above construction could be invented, and will enable the student to solve other analogous problems.

Analysis. Suppose that AB is divided in the required manner at the point H. Construct AFGH, the square on AH, and also the rectangle HB, BA, these being put on opposite sides of the line AB, as in the figure. Then it is natural to complete the square AB, and as FK is equal to AC, we see that DA is to be produced to F so that the rectangle contained by the whole line produced and the part produced may be equal to the square on AB. Thus the problem is reduced to a particular case of that considered in Prop B.

Ex. 1. If in the diagram to Prop. XI, CB and FG are produced to meet in R, shew that DHR is a straight line.

Ex. 2. Shew that, if the lines GB, FC and AK be drawn, they will all be parallel.

Ex. 3. If FC cut AB, HK in P, Q respectively; then FPQC.

Ex. 6. If KF cut AH in the point X, HX=BH.

Ex. 7. FB is 1' to DH.

Ex. 8. If DH and EB intersect in 0, 40 is parallel to FB, and perpendicular to DH.

Ex. 9. The square on EF is five times the square on EA.

Ex. 10. The sum of the squares on AB and BH is three times the quare on AH.

Ex. 11. The square on the sum of AB and BH is five times the square on AH.

Ex. 12. The difference of the squares on AH and HB is equal to the rectangle AH, HB.

Ex. 13. If, in the figure to Euclid II. 11, a point L be taken on ED produced such that EL-EB, and if a square ALMN be described so that the squares AC and AM are on opposite sides of ADL; shew that the line BA will be divided externally at N so that sq. on AN is equal to rectangle BA, BN.

Ex. 14. If X be taken on HA such that HX-HB, then square on IIX is equal to the rectangle HA, AX.

[This result is important. It shews that if a straight line be divided in medial section, and if the lesser segment be cut off from the greater, this latter is thereby divided in medial section. And this process can be continued.]

Ex. 15. DF is divided in medial section at A.

Ex. 18. Divide a straight line into two parts such that the sum of the squares on the whole line and one part may be equal to three times the square on the other part.

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Exercises in Euclid

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LECTURER ON GEOMETRY, ST. LUKE'S TRAINING COLLEGE, EXETER

London

MACMILLAN AND CO., LIMITED

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(a) Prove the corollary to this proposition.

(b) ABC is an isosceles triangle with AB equal to AC; the bisectors of the angles ABC and ACB meet at O. Prove that CO is equal to BO, and thence prove that the triangles AOB, AOC are equal in all respects.

(c) SPQR is a square, and PT, QT are drawn so as to make equal angles with PQ: show that ST is equal to TR.

(d) ABC is an isosceles triangle; equal parts AD, AE are cut off from the equal sides AB, AC respectively; BE, CD intersect at F: prove that FBC is an isosceles triangle.

(e) ABCD is a quadrilateral having AB equal to AD and each of the angles ABC, ADC a right angle. Prove that BC is equal to CD.

ON I. 7

(a) Can there be two equilateral triangles on the same base and on the same side of it?

(b) If two points be taken on the same side of a given straight line such that they are equidistant from a given point in that line, they cannot be equidistant from any other point in that line.

(c) Show that the beams of a roof keep the ridge in a fixed position.

(d) Two circles cannot intersect at more than one point on the same side of the line which joins their centres. (Employ the reductio ad absurdum method and suppose that they do.)

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FORMERLY SCHOLAR OF QUEEN'S COLLEGE, OXFORD

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"This Primer consists in the main of the First Course taken from the author's Elementary Mensuration. Rules have, however, been re-stated, and explanations re-written with a view to greater simplicity; moreover, such additions have been made in the Introductory Chapter as were requisite to bring the work within the reach of pupils with no previous knowledge of Geometry."

The schemes of work laid down by the Education Code for Day Schools and Evening Continuation Schools have been considered in determining the general standard of the book; but its scope is not limited to these requirements, the main purpose being to provide a practical introduction of the subject likely to be useful to all classes of beginners.

Some indication has been given of the service which the elements of Geometrical Drawing may render to Plane Mensuration.

The book may be studied, and the examples worked throughout, without the aid of Algebra, but for the convenience of those learners who have some knowledge of algebraical notation, many rules have been stated algebraically as well as verbally.

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