Similarity. Lemma In ΔABC, if E A B, F A C, and AE/AB = AF/AC, then E F B C.

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Lydia French
5 years ago
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1 Similarity Suppose l is a common transversal for lines m and m, all three lines lying in the same plane. Then l is parallel to neither m nor m, so we can set up a one-to-one correspondence between the points of m and those of m by associating to any point P on m the unique point P on m which is the intersection with m of the unique parallel l to l through P. This correspondence is called a parallel projection from m to m by the family of parallels to l. There are a number of properties enjoyed by the points on m which are preserved by their projection onto m. One is betweenness: if P-Q-R along m, then we must have P - Q - R along m also (why?). Another property preserved by projection is ratios of distances. In order to see this, we first prove a Lemma In ΔABC, if E A B, F A C, and then E F B C. AE/AB = AF/AC, Proof Suppose AE/AB = AF/AC but E F is not parallel to B C. Then the parallel to B C through F meets A B in a point G E and the the parallel to B C through E meets A C in a point H F.

2 Moreover, if A-G-E then A-F-H, while if A-E-G then A-H-F; as both cases are similar, we shall assume the latter. Next, we repeatedly subdivide segments A B and A C by adding first their midpoints, then the midpoints of the two subdivided segments so produced, then the midpoints of those subdivided segments, etc., until one of these midpoints M lies between E and G. (This is possible since the lengths of the subdivided segments are successively AB / 2, AB / 2 2, AB / 2 3,, and eventually these lengths become smaller than EG.) Since at every stage of this process, the segment joining corresponding midpoints on A B and A C is parallel to B C (by the Midpoint Connector Theorem either for Triangles or Trapezoids), the segment M N joining M to the corresponding midpoint N on A C is parallel to B C and has H-N-F (as projection preserves betweenness). If midpoints M and N were produced at the nth stage of this process and, counting down from A, M is the kth of these midpoints along A B and the distance between each consecutive pair is b, then, also counting down from A, N is the kth of the midpoints along A C, and the distance between each consecutive pair here is equal to some constant c. Thus AB = nb, AC = nc, and AM = kb, AN = kc. So AM/AB = k/n = AN/AC, and since AE < AM while AF > AN, we have AE/AB < AM/AB = AN/AC < AF/AC, contradicting our hypothesis. //

3 Theorem [The Side-Splitting Theorem] If a line parallel to side B C of ΔABC cuts A B in E and A C in F, then AE/AB = AF/AC and AE/EB = AF/FC. Proof If the hypotheses held but AE/AB AF/AC, then since 0 < AX/AB < 1 as X runs between A and B and 0 < AF/AB < 1, there must (by the Ruler Postulate) be some point E E on A B so that A E /AB = AF/AC. By the lemma, then E F B C. But E F B C, contradiction. So AE/AB = AF/AC. Cross multiplying gives AE AC = AF AB AE (AF + FC) = AF ( AE + EB) AE AF + AE AC = AF AE + AF EB AE AC = AF EB whence AE/EB = AF/FC. // Corollary [Parallel Projection Preserves Ratios of Segments] If l is a common transversal for lines m and m and P-Q-R along m (so that P - Q - R along m for the parallel projection points), then PQ/QR = P Q / Q R.

4 Proof If m m, then PQ Q P and QR R Q are parallelograms, so PQ = P Q and QR = Q R and the result is trivial. If m meets m at O, then there are a number of cases depending on where O lies relative to P-Q-R along m. If O-P-Q-R, then by the Side-Splitting Theorem, OP/PQ = O P / P Q, OP/OQ = O P /O Q and OQ/QR = O Q / Q R. Multiply the second and third of these proportions and divide by the first to find PQ/QR = P Q / Q R (and by algebra similar to the lines at the end of the proof of the theorem above, we also have PQ/PR = P Q / P R ). If P-Q-R-O, then the same argument applies with P, P switching places with R, R ; but this produces the reciprocal of the desired proportion, so the result holds here as well. If O lies between P and Q, then find X so that O is the midpoint of PX and X so that O is the midpoint of P X ; now, since OP = OX and O P = O X, then O-X-Q-R, O-Q-X-R or O-Q-R-X, and in each case, the desired proportion follows from the argument in the first case above. //

5 Two triangles ΔABC and ΔXYZ are said to be similar, denoted ΔABC ~ ΔXYZ, if there is some one-to-one correspondence of their vertices so that the induced correspondence on their angles matches congruent angles, and the induced correspondence on their sides matches sides with constant ratios, the value of which is called the constant of proportionality, or scale factor, of the similarity. Theorem If a line parallel to side B C of ΔABC cuts A B in E and A C in F, then AE/AB = AF/AC = EF/BC. // Corollary [AA similarity] If ΔABC and ΔXYZ are triangles for which ABC XYZ and BCA YZX, then ΔABC ~ ΔXYZ. // Corollary [SAS similarity] If ΔABC and ΔXYZ are triangles for which ABC XYZ and AB / XY = BC / YZ, then ΔABC ~ ΔXYZ. // Corollary [SSS similarity] If ΔABC and ΔXYZ are triangles for which AB/ XY = BC/YZ = CA/ZX, then ΔABC ~ ΔXYZ. //

UNIT 1: SIMILARITY, CONGRUENCE, AND PROOFS. 1) Figure A'B'C'D'F' is a dilation of figure ABCDF by a scale factor of 1. 2 centered at ( 4, 1).

UNIT 1: SIMILARITY, CONGRUENCE, AND PROOFS. 1) Figure A'B'C'D'F' is a dilation of figure ABCDF by a scale factor of 1. 2 centered at ( 4, 1). 1) Figure A'B'C'D'F' is a dilation of figure ABCDF by a scale factor of 1. 2 centered at ( 4, 1). The dilation is Which statement is true? A. B. C. D. AB B' C' A' B' BC AB BC A' B' B' C' AB BC A' B' D'