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Make the angle ABE equal to the angle DBC: Then, because the angle ABE is equal to the angle DBC, and the angle BAE to the angle BDC in the same segment, the triangle ABE is equiangular to the triangle DBC: Therefore BA is to AE as BD to DC, and therefore the rectangle AB, CD is equal to the rectangle BD, AE: Again, because the angle ABE is equal to the angle DBC, to each of these add the angle EBD; therefore the angle ABD is equal to the angle EBC: And the angle BDA is equal to the angle BCE in the same segment; therefore the triangle ABD is equiangular to the triangle EBC: Therefore AD is to DB as EC to CB, and therefore the rectangle AD, BC is equal to the rectangle BD, EC: But the rectangle AB, CD has been proved equal to the rectangle BD, AE; therefore (2. 1) the whole rectangle BD, AC is equal to the two rectangles AB, CD and AD, BC.
Wherefore, The rectangle &c. Q.E.D.
1. The rectangle contained by two lines is a mean proportional between their squares.
2. A quadrilateral is such that the perpendiculars upon a diagonal from opposite angles are equal : divide it into four equal triangles by lines drawn to the angles from a point within it.
3. From one angle of a triangle a line is drawn to the bisection of the opposite side, and through its middle point another is drawn from either angle to the side subtending it: prove that this side is divided in the ratio of 2 to 1.
4. Through a given point draw a line so that, if perpendiculars be dropped upon it from two other given points, the parts intercepted between the given point and the feet of the perpendiculars may be equal.
5. If any three equal lines be drawn from the angles of a triangle to the opposite sides or these produced, and from any point within it lines parallel to these be drawn to the sides, the sum of these latter lines shall be equal to either of the former.
6. If three circles touch each other, two of which are equal, the vertical angle of the triangle, formed by joining the points of contact, is equal to the angle at the base of the isosceles triangle, formed by joining the three centres.
7. ABC is an equilateral triangle, E any point in AC; in BC produced take CD, CF equal to CA, CE, respectively, and let AF, DE, intersect in H: shew that
HC:EC :: AC:AC+EC,
8. The diagonals, AC, BD, of an inscribed quadrilateral meet in E: shew that AB.BC: AD.DC::BE:ED.
9. If a square be inscribed in a right-angled triangle, one side coinciding with the hypothenuse, the base is divided in continued proportion.
10. If ABC be an inscribed triangle, and BD be drawn parallel to the tangent at A to meet AC, or AC produced, shew that AB is a mean proportional between AC, AD.
11. AB is divided in C, D, so that AB:AC::AC:AD; if AE be any other line taken equal to AC, shew that the angle BED is bisected by EC.
12. The part of a tangent to a circle, intercepted by tangents at the extremities of any diameter, is divided at the point of contact so that the radius is a mean proportional between the two segments.
13. If two chords in a circle intersect so that the segments of one have the same ratio as those of the other, the line bisecting the angle between homologous segments will pass through the centre of the circle.
14. If a triangle be inscribed in a semicircle, and a perpendicular be drawn from any point in the diameter, meeting the circumference and the other two sides, the three segments of it thus made will be in continued proportion.
15. If the diagonals of an inscribed quadrilateral intersect at right angles, the rectangles of the opposite sides are together double of the area of the figure.
16. If chords, drawn from any fixed point in the circumference of a circle, be cut by another chord, parallel to the tangent at that point, the rectangle of each chord and the part of it between the given point and given chord is constant.
17. If in similar triangles, from any two equal angles, lines be drawn to the opposite sides, making equal angles with homologous sides, these will have the same ratio as the sides on which they fall, and will also divide them proportionally.
18. Apply Prop. 11. to shew how a line may be drawn through a given point, parallel to a given line, by means of a piece of string.
19. In any triangle, right-angled at A, if CD be drawn bisecting the angle C, shew that
AB : AC :: BC-AC: AD. 20. If two circles touch externally, the part of their common tangent between the points of contact is a mean proportional between the diameters.
21. Given the lengths of three lines, drawn from the angles to the points of bisection of the opposite sides, to construct the triangle.
22. Inscribe a square in a given segment of a circle.
23. Find the locus of points dividing in a given ratio lines drawn from a given point to meet a given circle.
24. Given two circles which intersect, draw through either point of intersection a line cutting the circles, so that the chords intercepted may be in a given ratio.
25. Inscribe in a given triangle a parallelogram similar to a given parallelogram.
26. If through the vertex and extremities of the base of a triangle any two circles be described so as to intersect in the base or base produced, their diameters will be proportional to the respective sides.
27. If any two lines which intersect are cut by two given lines, the ratio of the rectangles of their respective segments will be the same as that for any other two lines, drawn parallel to them, and cut by the same given lines.
28. A, B, are the centres of two unequal circles, AP, BQ, any pair parallel radii : shew that PQ passes through a fixed point, whose distances from the centres are proportional to the respective radii ; and hence deduce a method of drawing a common tangent to two circles.
29. If through the fixed point in  any line be drawn cutting the circles, the radii at the points of section will be parallel, and the intercepted chords proportional to them.
30. If through the fixed point in  any two lines be drawn cutting the circles, the chords joining corresponding points of section will be parallel ; and, of the eight points of section, any four lie in the circumference of a circle, provided that no two of them be homologous points, nor at once in the same circle and line.
31. CDE is a common tangent to two circles, meeting the line of centres, AB, in E; FGHKE is
line cutting the circles : shew that
EC.ED=EF.EK=EG.EH. 32. If any circle be drawn touching two other given circles, the lines joining the points of contact will pass through a fixed point.
33. Find the Arithmetic, Geometric, and Harmonic means between three given lines. *
• Three straight lines AB, AC, AD, are said to be (i) in Arith. Prog., when AB: AB :: AB~ AC: AC~AD, (ii) in Geom. Prog., when AB: AC :: AB~ AC: AC ~AD, (iii) in Harm. Prog. when AB:AD :: AB ~ AC: AC~ AD. Hence it will be found that if AP, AQ, AR, be, respectively, the Ar., Geom., and Harm, means between AB and AC,
2AB.AC AP = }(AB + AC), AQ* = AB.AC, AR =
AB+AC' whence also AP.AR = AQ’, or the three means are in continued proportion.
The Proportion in (iii) is often thus expressed, AB:AD::CB:CD, or AD:AB::CD:CB or AD.BC=AB.CD.