Geometry. figure (e.g. multilateral ABCDEF) into the figure A B C D E F is called homothety, or similarity transformation.

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1 ctober 18, 2015 Geometry. Similarity and homothety. Theorems and problems. efinition. Two figures are homothetic with respect to a point, if for each point of one figure there is a corresponding point belonging to the other figure, such that lies on the line at a distance from point, and vice versa, for each point of the second figure there is a corresponding point belonging to the first figure, such that lies on the line F Homothety F at a distance from point. Here the positive number is called the homothety (or similarity) coefficient. Homothetic figures are similar. The transformation of one figure (e.g. multilateral F) into the figure F is called homothety, or similarity transformation. Thales Theorem orollary 1. The corresponding segments (e.g. sides) of the homothetic figures are parallel. Thales Theorem orollary 2. The ratio of the corresponding elements (e.g. sides) of the homothetic figures equals. xercise. What is the ratio of the areas of two similar (homothetic) figures? efinition. onsider triangles, or polygons, such that angles of one of them are congruent to the respective angles of the other(s). Sides which are adjacent to the congruent angles are called homologous. In triangles, sides opposite to the congruent angles are also homologous.

2 Generalized Pythagorean Theorem 2. Theorem 2. For three homologous segments,, and belonging to the similar right triangles, and, where is the altitude of the triangle drawn to its hypotenuse, the following holds, l l l Proof. If we square the similarity relation for the homologous segments,, where, and are the legs and the hypotenuse of the triangle, we obtain,. Using the property of a proportion, we may then write, Pythagorean theorem for the right triangle,, we immediately obtain., wherefrom, by Theorem 1. If three similar polygons, P, Q and R with areas, and are constructed on legs, and hypotenuse, respectively, of a right triangle, then, Q Proof. The areas of similar polygons on the sides of a P a b a c right triangle satisfy and, or,. Using the property of a proportion, we R may then write,, wherefrom, using the Pythagorean theorem for the right triangle,, we immediately obtain.

3 Property of the bisector. Theorem (property of the bisector). The bisector of any angle of a triangle divides the opposite side into parts proportional to the adjacent sides,,, Proof. onsider the bisector. raw line parallel to from the vertex, which intercepts the extension of the side at a point. ngles and have parallel sides and therefore are congruent. Similarly are congruent and. Hence, triangle is isosceles, and =. Now, applying the intercept theorem to the triangles and, we obtain. Theorem (property of the external bisector). The bisector of the exterior angle of a triangle intercepts the opposite side at a point ( in the Figure) such that the distances from this point to the vertices of the triangle belonging to the same line are proportional to the lateral sides of the triangle. Proof. raw line parallel to from the vertex, which intercepts the side at a point. ngles and have parallel sides and therefore are congruent. Similarly, we see that angles and are congruent, and, therefore,. pplying the intercept theorem, we obtain,.

4 Selected problems on similar triangles (from last homework). Problem 1 (5). In the isosceles triangle point divides the side into segments such that. If H is the altitude of the triangle and point is the intersection of and, find the ratio to. 2 Solution. First, let us perform a supplementary construction by drawing the segment parallel to,, where point belongs to the side, and point to and the altitude. Notice the similar triangles,, which implies,. y Thales theorem,, and, so 1 F H that., because. Therefore, the sought ratio is,. Problem 2 (6). Point belongs to the continuation of side of the triangle such that =. Point F belongs to side, and F = 3 F. Segment F intercepts side at point. Find the ratio :. Solution. First, let us perform a supplementary construction by drawing the segment parallel to,, where belongs to the side of the triangle. is the mid-line of the triangle, and, by Thales, also of and. Therefore,, and, so. n the other hand, again, by Thales, or, noting similar triangles,. G F

5 ommensurate and incommensurate segments. The uclidean algorithm. efinition. Two segments, and, are commensurate if there exists a third segment,, such that it is contained in each of the first two segments a whole numbers of times with no remainder.,,, s b=ms s a=ns The segment is called a common measure of the segments and. The concept of commensurability is similar to that of the common divisor for integers. It can be extended to any two quantities of the same denomination two angles, two arcs of the same radius, or two weights. The greatest common measure. If a common measure of two segments and is sub-divided into two, three, or, generally, any number of equal smaller segments, these smaller segments are also common measures of the segments and. In this way, an infinite set of common measures, decreasing in length, can be obtained,,. Since any common measure is less than the smaller segment,, there must be the largest among the common measures, which is called the greatest common measure. Finding the greatest common measure (GM) is done by the method of consecutive exhaustion called uclidean algorithm. It is similar to the method of consecutive division used for finding the greatest common divisor in arithmetic. The method is based on the following theorem. Theorem. Two segments and are commensurate, if and only if the smaller segment,, is contained in the greater one a whole number of times with no remainder, or with a remainder,, which is commensurate with the smaller segment,.,,.

6 The greatest common measure of two segments is also the greatest common measure of the smaller segment and the remainder, or there is no remainder. Proof. First, consider the necessary condition. Let and be commensurate,,,, and. Let be their greatest common measure. Then, either ( ) and segment is contained in a whole number of times with no remainder, being the GM of the two segments, or,,. Then,, where, and, therefore,,, which shows that and are commensurate. The sufficiency follows from the observations that (i) if segment is contained in a whole number of times with no remainder, then the segments are commensurate, and is the greatest common measure of the two, while (ii) if, and and are commensurate with the greatest common measure,,, then,, and and are also commensurate with the same GM. The uclidean algorithm. In order to find the GM of the two segments, and, we can proceed as follows. First, using a compass exhaust the greater segment, marking on it the smaller segment as many times as possible, until the remainder is smaller than the smaller segment,, or there is no remainder. ccording to rchimedes exhaustion axiom, these are the only two possible outcomes. Following the above theorem, the problem now reduces to finding the GM of this remainder,, and the smaller segment,. We now repeat the same procedure, exhausting segment with, and again, there is either no remainder and is the GM of and, or there is a remainder. The problem is then reduced to finding the GM of a pair of even smaller segments, and, and so on. If segments and are commensurate and their GM,, exists, then this process will end after some number of steps, namely, on step where. Indeed, all remainders in this process are multiples of,, and is the decreasing sequence of natural numbers, which necessarily terminates, since any non-empty set of positive integers has the smallest number principle of

7 the smallest integer. If the procedure never terminates, then segments have no common measure and are incommensurate. and xample. The hypotenuse of an isosceles right triangle is incommensurate to its leg. r, equivalently, the diagonal of a square is incommensurate to its side. Proof. onsider the isosceles right triangle shown in the Figure. ecause the hypotenuse is less than twice the leg by the triangle inequality, the leg can only fit once in the hypotenuse, this is marked by the segment. Let the perpendicular to the hypotenuse at point intercept leg at point. Triangle is also isosceles. This is because angles and supplement equal angles and to 90 degrees, and therefore are also equal. Triangle is an isosceles right triangle, similar to. Its leg = - = = is a remainder of subtracting the leg = from the hypotenuse,, while the hypotenuse, = - = -, is the remainder of subtracting this remainder from the leg =. Hence, on the second step of the uclidean algorithm we arrive at the same problem as the initial one, only scaled down by some overall factor. bviously, this process never ends, and therefore the hypotenuse and the leg are incommensurate.