20. Shew that IV. 14 may be deduced from IV. 18.

21. ABC is a triangle, AD the perpendicular on BC, AOE the diameter of the circle about ABC. Shew that

BA.AC=EA.AD.

22. ABCD is a quadrilateral figure in a circle. Shew that

AC.BD=AB. CD+BC.DA.

Make the angle BAE= CAD. The triangles ABE, ACD are similar, as also the triangles ADE, ACB.

Therefore

AB: RE=AC: CD and AB, CD = BE. AC,

AD: RE=AC: CB and AD.CB=DE.AC.

2A 4Pisarriangle, 47 bisects and meets the base la 7 and the circumscribed circle in E; BAD, EAC are similar mängis and

CONSTRUCTIONS.

1. To trisect a given line.

2. To divide a given line into 5 parts, or any number of parts.

3. To find a fourth proportional to three given straight lines.

4. To find a third proportional to two given straight lines.

5. To find a mean proportional to two given straight lines.

Place them in one straight line, on this describe a semicircle, and at the point where the lines join erect a perpendicular to meet the circumference. This will be the mean proportional required.

Construct a square equal to a given rectangle.

[Iv. 12.

6. To divide a line into three parts which are to one another as 3 5 7 or as m: n: p.

7. To divide a line AB in extreme and mean ratio, i. e. so that AB : AC=AC : CB or AB. BC = AC2.

On AB describe a square ABDE, bisect EA in F. Join FB, and produce FA to G, making FG = FB. On AG describe a square: the angle of this square determines the required point on AB.

The proof from III. 11 and III. 12.

8. To construct an isosceles triangle having each of the angles at the base double of the third angle.

Take any straight line AB, and divide it so that
AB.BC=AC2.

With centre A and distance AB describe a circle, in

which place BD = AC. Join AD; ABD is the triangle

required. Join DC. ABD, DBC are similar triangles IV. 4. And therefore DC = DB = AC, whence CAD CDA, and ABD=DCB= twice CAD.

=

9. To inscribe a regular pentagon in a given circle.

The inscription of a regular figure in a circle depends upon our being able to divide the circumference into as many equal parts as the figure has sides.

We have seen how to divide a circumference into 3 parts or 4 parts, and since every arc can be bisected, we can divide a circumference into 6, 12, 24, &c. or 8, 16, 32, &c. parts.

The preceding construction gives us an angle CDB equal to one-fifth of two right angles. If therefore DC be produced to meet the circumference, it will intercept one-fifth part of it, Complete the construction.

10. To inscribe a regular decagon in a given circle.

11. To inscribe a regular quindecagon in a given circle.

12. On a given straight line to construct a polygon similar to a given polygon.

13. To inscribe a square in a triangle.

14. To inscribe a square in a given segment of a circle. 15. To describe a circle touching two straight lines and passing through a fixed point.

16. To describe a circle touching a straight line and passing through two fixed points.

17. To divide a triangle into two parts in the ratio of two given lines by a line drawn parallel to its base.

18. To produce a line AB to C so that AB, AC may be equal to a given square.

19. To construct a rectangle equal to a given square, and such that the difference of its sides may be equal to a given straight line.

20. To construct a figure similar to either of two given similar figures, and equal to their sum or their difference.

21. To construct a figure equal to one and similar to another given figure.

1. O is a point within the angle ACB. To draw through O a straight line AOB, so that it may be bisected in 0.

2. A and B are two points on the same side of the straight line CD. Find in CD a point P such that PA+PB may be the least possible.

3. In the figure of III. 12 shew that AL, CF, BK pass through one point.

4. An equilateral triangle is inscribed in a circle, if the intercepted arcs be bisected and the points of bisection joined, the sides of the triangles are trisected,

5. Through four given points to draw lines forming a

square.

6. Shew that the circumference of a circle is more than three times and less than four times as great as the diameter.

7. Draw from the vertex of a triangle a line to the base such that it may be a mean proportional between the segments

of the base.

8. AB is the diameter of a circle, DE a chord, OP a perpendicular on AB from any point 0 of DE. Shew that AP.PB-DO. OE+ OP2.

9. The part of a variable tangent intercepted between two fixed tangents subtends a constant angle at the centre.

10. ABC is a triangle. AD, EE, CF are lines drawn through one point to meet the opposite sides. Shew that AF.BD.CE=FB.DC.EA.

11. The common tangents to two circles meet in a point 0; and through a chord OKGMN is drawn meeting the cinumferences in A, G, M and M. Shew that

OK. ON=0G. OM

12. Given the perpendiculars to constract the triangle.

CAMBRIDGE: PRINTET AT TEE TNIVERSITY PRESS.