and from B draw BD at right angles to AC, and meeting the circumference at D :

then shall BD be a mean proportional between AB and BC.

Join AD, CD.

Because ADC is a semicircle,

therefore the angle ADC is a right angle;

and DB is the perpendicular to the hypotenuse of the right-angled

triangle DAC,

therefore AB: BD :: BD : BC,

that is, BD is a mean proportional between AB and BC.

V. 7, Cor.

Q.E.F.

PROB. 6. On a given straight line to describe a rectilineal figure similar and similarly situated to a given rectilineal figure on another given straight line.

Let AB be the given straight line, CDEFG the given rectilineal figure :

it is required to describe on AB a rectilineal figure similar and similarly situated to CDEFG on CD.

K

H

E

B

Divide the figure CDEFG into triangles by joining C to E and F.

Make the angles BAH, ABH equal to the angles DCE, CDE respectively,

and make the angles HAK, AHK equal to the angles ECF, CEF respectively,

and so on until triangles have been made equiangular to all the triangles into which the given figure was divided :

then shall ABHKL be the rectilineal figure required.

By construction, the figure ABHKL is equiangular to the figure CDEFG.

Because the triangles ABH, CDE are equiangular, therefore AB BH :: CD: DE,

and BH: HA :: DE: EC ;

and because the triangles AHK, CEF are equiangular,

therefore HA: HK :: EC: EF;

therefore BH: HK::DE : EF,

In like manner HK : KL :: EF: FG,

and so on.

V. 2

Ex æquali.

Therefore the sides about the equal angles of the figures are proportional.

Hence the figure ABHKL is similar to the figure CDEFG.

Q.E.F.

Ex. 60. Of all the figures that can be described on the given line similar to the given figure, the greatest has to the least the duplicate ratio of the greatest side of the figure to the least.

PROB. 7. To describe a rectilineal figure equal to one and similar to another given rectilineal figure.

Let E and S be two given rectilineal figures :

it is required to describe a rectilineal figure equal to the figure E and similar to the figure S.

On AB a side of S construct the rectangle ABCD equal to the figure S, II. Prob. 3. and on BC construct the rectangle BCFH equal to the figure E. Take KL a mean proportional between AB and BH, and on KL describe the rectilineal figure T similar to S, so that KL and AB are homologous sides of T and S: V. Prob. 6

then shall T be the rectilineal figure required.

Because AB KL:: KL: BH,

therefore AB is to

:

BH in the duplicate ratio of AB to KL, but the figure S is to the figure T in the duplicate ratio of AB to KL, therefore the figure S is to the figure T as AB is to BH.

And because rectangles of equal altitude are to one another as their bases,

therefore the rectangle AC is to the rectangle BF as AB is to BH. Therefore the figure S is to the figure T as the rectangle AC is to the rectangle BF.

But the figure S is equal to the rectangle AC,

therefore the figure T is equal to the rectangle BF,

and therefore to the given figure E.

Therefore T is the rectilineal figure required.

Q.E.F.

EXERCISES.

61. Through a given point A draw a line meeting the given lines OX, OY in points P, Q respectively, so that OP is to OQ in a given ratio.

62. Through two fixed points in the circumference of a circle draw two parallel chords which shall be to each other in a given ratio.

63. Construct a triangle similar to a given triangle, having one vertex fixed and the others on two given straight lines.

64. Draw from a given point P two straight lines PQ, PR at a given inclination to one another to meet two given straight lines in Q and R so that PQ,PR may be equal.

65. Describe an isosceles triangle equal to a given triangle and having a given vertical angle.

66. Three straight lines meet in a point. cutting them so that the intercepted segments magnitudes.

Draw a fourth may have given

67. In a given square inscribe a square of which the area shall be equal to three-fourths of that of the given square.

EXERCISES ON BOOK V.

68. The perpendiculars from the vertices of a triangle upon the opposite sides are inversely proportional to the sides to which they are respectively perpendicular.

69. If in the same circle there be inscribed two triangles of equal area, then the rectangle contained by any two sides of the one is to the rectangle contained by any two sides of the other, as the base of the second is to the base of the first.

70. Produce a given straight line so that the rectangle contained by it and the produced part shall be equal to a given square.

71. From the same point A straight lines are drawn, making the angles BAC, CAD, DAE equal to one another, and they are cut by a straight line BCDE, which makes BAE an isosceles triangle: shew that BC or DE is a mean proportional to BE and CD.

72. Given the base of a triangle, and the point where the line bisecting the exterior vertical angle cuts the base produced, find the locus of the vertex of the triangle.

73. P is any point in the circumference of a circle whose centre is O. Join OP, and produce it to Q, making PQ equal to n times OP, and from Q draw a straight line touching the circle in R. Join PR, and shew that the diameter of the circle described about the triangle PQR is equal to (n + 1) PR.

74. ABD is the diameter of a semicircle ACD, and ABC is a right angle. E, any point on the chord AC within the semicircle, is joined to B, and CF is drawn cutting AD in F,