Then, by like reasoning, it may be proved that AH, BK, CL, DM cut one another internally in the same point O in the ratio of homologous sides of the figures. In this case the proof holds when the figures are equal, but we may then obtain the result more simply from the theorem that the diagonals of a parallelogram bisect one another. Again, let any straight line be drawn through O to meet corresponding sides AB, HK in P, Q respectively: then shall OP be to OQ as AB is to HK. Because AP and HQ are parallel, therefore the triangles APO, HQO are equiangular and therefore OP : OQ :: OA : OH; V. 2. but OA: OH :: AB: HK, therefore OP: OQ:: AB: HK. Q.E.D. COR. Similar rectilineal figures may be divided into the same number of similar triangles. For if the figures be placed so that their corresponding sides are parallel, and one figure lies wholly within the other, the lines joining corresponding angular points will meet in a point within the figures, and will divide the figures into as many pairs of similar triangles as each figure has sides. DEF. 4. The point of intersection of all straight lines which join the corresponding points of two similar figures, whose corresponding sides are parallel, is called the centre of similarity of the two figures. Ex. 10. If any point O is joined to the angular points A, B, C... of a rectilineal figure, and points a, b, c are taken in OA, OB, OC... such that Oa: OA :: Ob : OB :: Oc : OC, etc., then the figure abc... is similar to the figure ABC... Ex. 10a. If ABCD ..., abcd . . . are two figures such that the lines Aa, Bb, Cc... meet in the same point O, then, if the sides AB, BC ... are respectively parallel to ab, bc.. the figures are similar. Is it true that, if the figures are similar, the sides are parallel? Ex. II. O is a fixed point, P any point on the circumference of a given circle. Q is taken on OP such that OQ: OP is a fixed ratio. Prove that Q lies on the circumference of a circle. Ex. 12. O is a fixed point, P any point on the circumference of a given circle. OQ is taken making a given angle with OP, and having a fixed ratio to it. Shew that Q lies on the circumference of one of two circles. THEOR. 7. In a right-angled triangle, if perpendicular is drawn from the right angle to the hypotenuse, it divides the triangle into two other triangles which are similar to the whole and to one another. Let ABC be a triangle having the angle BAC a right angle, and let AD be perpendicular to BC: then shall the triangles DBA, DAC be similar to the triangle ABC and to one another. In the triangles DBA, ABC, the angle ABC is common, and the right angles BDA, BAC are equal, therefore the triangles are similar. V. 2. In like manner the triangles DAC, ABC are similar. Hence the triangles DBA, DAC being similar to the same triangle ABC are also similar to one another. V. 1. Q.E.D. COR. Each side of the triangle is a mean proportional between the hypotenuse and the adjacent segment of the hypotenuse; and the perpendicular is a mean proportional between the segments of the hypotenuse. Ex. 13. Name the sides of the three triangles which are homologous to one another. Ex. 14. The segments of the base of the triangle are in the duplicate ratio of the sides of the triangle. Ex. 15. Shew that the radius of a circle is a mean proportional between the segments of any tangent between the point of contact and any two parallel tangents. Ex. 16. If two circles touch externally, the part of each of the two symmetrical common tangents between its points of contact is a mean proportional between the diameters. THEOR. 8. If from any angle of a triangle a straight line is drawn perpendicular to the base, the diameter of the circle circumscribing the triangle is the fourth proportional to the perpendicular and the sides of the triangle which contain that angle. Let AE be a diameter of the circle circumscribing the triangle ABC, AD the perpendicular on BC: then shall AD be to AC as AB to AE. Join BE. Because ABE is a semicircle, therefore the angle ABE is a right angle, therefore the angle ADC is equal to the angle ABE. III. 17. Because the angles ACD and AEB are in the same segment ACB, therefore the angle ACD is equal to the angle AEB. Hence the triangles ACD, AEB are equiangular, and therefore AD: AC:: AB: AE. III. 16. V. 2. Q.E.D. Ex. 17. D is a point on the base BC, or base produced, of the isosceles triangle ABC: shew that the circumscribed circles of the triangles ABD, ACD are equal. Ex. 18. D is any point on the side BC of the triangle ABC: shew that the diameters of the circles about the triangles ABD, ACD are in the ratio of AB to AC. THEOR. 9. If the interior or exterior vertical angle of a triangle is bisected by a straight line which also cuts the base, the base is divided internally or externally in the ratio of the sides of the triangle. And, conversely, if the base is divided internally or externally in the ratio of the sides of the triangle, the straight line drawn from the point of division to the vertex bisects the interior or exterior vertical angle. Let ABC be a triangle, and let CP and CQ, the bisectors of the interior and exterior vertical angles at C, cut the base AB in P and Q: P then shall AB be divided internally at P and externally at Q in the ratio AC: CB. |