Make the angle BAE equal to the angle CAD, and make the angle ABE equal to the angle ACD. Join DE. Then the triangles ABE, ACD are equiangular, therefore BA : AE :: CA : AD, therefore BA : AC :: EA : AD; but the angle BAE is equal to the angle CAD, V. 2. Alternando. therefore the whole angle BAC is equal to the whole angle EAD; therefore the triangles BAC, EAD are similar. Because the triangles ABE, ACD are similar, therefore AB BE:: AC: CD, V. 3. therefore the rectangle contained by AB and CD is equal to the rectangle contained by AC and BE. Again, because the triangles BAC, EAD are similar, therefore BC: AC :: DE: AD, therefore the rectangle contained by BC and AD is equal to the rectangle contained by. AC and DE. Hence the sum of the rectangles contained by AB and CD and by BC and AD is equal to the sum of the rectangles contained by AC and BE and by AC and DE. But because BD is less than the sum BE and ED, therefore the rectangle contained by AC and BD is less than the sum of the rectangles contained by AC and BE and by AC and DE, therefore the rectangle contained by AC and BD is less than the sum of the rectangles contained by AB and CD and by BC and AD. But if the quadrilateral ABCD be such that a circle can be described about it, then the angle ABD is itself equal to the angle ACD, therefore the point E lies on BD, and therefore BD is equal to the sum of BE and ED, therefore the rectangle contained by AC and BD is equal to the sum of the rectangles contained by AB and CD and by BC and AD. Q.E.D. Ex. 46. If the diagonals of a quadrilateral inscribed in a circle cut each other at right angles, shew that the rectangles contained by the opposite sides are together double of the quadrilateral figure. EXERCISES. 47. If ACB, BCD be equal angles and DB be perpendicular to BC and BA to AC, prove that the triangle DBC is to the triangle ABC as DC is to CA. 48. ABCD is a square, P a point on the arc AB of its circumscribing circle. Shew that the rectangle PC, PD is equal to the sum of the rectangles PA, PB; PB, PC; PD, PA. 49. CA, CB are two radii of a circle at right angles to each other, DE is any chord. If BD, BE meet AC in F and G, prove that the triangles BFG, BDE are similar. 50. Similar triangles are in the ratio of the squares on the radii of their inscribed (or circumscribed) circles. 51. The sides of a regular hexagon are produced both ways, and the points of intersection joined to form a new regular hexagon. Prove that the area of the second hexagon is to that of the first in the ratio 3 : 1. 52. If from a point without a circle two straight lines be drawn, one of which touches and the other cuts the circle, a line drawn from the same point in any direction equal to the tangent will be parallel to the chord of the arc intercepted by two lines drawn from its other extremity to the intersections of the circle with the cutting line. 53. If ABC be an equilateral triangle, and P a point on the circumscribing circle on the side of BC remote from A, prove that the square on PA is equal to the rectangle PB, PC together with the square on BC. 54. On the side BC of an acute-angled triangle ABC as diameter a circle is described. On AB a point D is taken such that AD is equal to the tangent drawn from A to the circle, and DE is drawn at right angles to AB to meet AC produced in E. Prove that the triangle ADE is equal to the triangle ABC. 55. On the sides of a triangle ABC, points D,E,F are taken, so that AD: DB:: BE:EC::CF: FA :: 1 : 2. Find the ratio of the triangle DEF to the triangle ABC. *56. If the vertical angle of a triangle be bisected internally or externally by a straight line which meets the base, the square on the part of the bisector between the vertex and the base is equal to the difference of the rectangle contained by the sides of the triangle and that contained by the segments of the base. SECTION III. Loci AND PROBLEMS. LOCUS I. To find the locus of a point whose distances from two intersecting straight lines are in a given ratio. Let AB and CD be two straight lines intersecting at O; it is required to find the locus of a point whose distances from AB and CD are in a given ratio. Let P be any point such that PM the perpendicular on AB is to PN the perpendicular on CD in the given ratio. From P draw PH parallel to ON to meet AB at H, and from P draw PK parallel to OM to meet CD at K. Let P lie within the angle BOD or the vertically opposite angle, then the angle PHM is equal to the angle PKN, because each is equal to the angle BOD, and the right angles PMH, PNK are equal, therefore the triangles PMH, PNK are similar, V. 2. therefore PH: PM :: PK: PN, therefore PH: PK :: PM: PN, that is PH: PK is equal to a given ratio, or OK: PK is a given ratio; and the angle OKP is constant, because it is supplementary to BOD, therefore the angle POK is constant, therefore P lies on a fixed straight line through O. Alternando. V. 3. In like manner it may be shown that if P' be a point within the angle BOC or the vertically opposite angle that P' lies on a second fixed straight line through O. Also the distances of every point on these two straight lines from AB and CD are in the given ratio. Let Q be such a point, draw QR perpendicular to AB, and QS perpendicular to CD. Then the triangles QRO, PMO are equiangular, therefore QR: PM :: OQ: OP, and the triangles QSO, PNO are equiangular, therefore QS PN:: OQ: OP, : therefore QR: PM :: QS: PN, V. 2. V. 2. therefore QR: QS :: PM : PN, Alternando. therefore the distances of Q from AB and CD are to one another in the given ratio. |