Parallel Projection Theorem (Midpoint Connector Theorem):

Transcription

1 rllel rojection Theorem (Midpoint onnector Theorem): The segment joining the midpoints of two sides of tringle is prllel to the third side nd hs length one-hlf the third side. onversely, If line isects one side of tringle nd is prllel to the second, it lso isects the third side. ~ egin with ª with L nd M the midpoints of nd respectively. xtend LM to find with L*M* nd LM = M. onnect nd. y construction nd SS, ªLM ªM nd so (pl p, lternte interior ngles). Note lso tht L = L =. If we construct L, we hve pl pl (lternte interior ngles with L s the 1 1 trnsversl) so we get ªL ªL y SS. Thus LM = L = y, nd since 2 2 pl pl, nd they re lternte interior ngles with L s the trnsversl, LM. The converse follows from uniqueness of prllels. If line isects one side of the tringle, is prllel to second side, ut does not isect the third, drw the line connecting the two midpoints. We just proved tht line must e prllel to the se, ut it contrdicts the uniqueness of prllels through the midpoint of the first side. L M

2 Definition: medin of trpezoid is line segment whose endpoints re the midpoints of the two legs. Theorem (Midpoint onnector Theorem for Trpezoids): If line isects one leg of trpezoid nd is prllel to the se, then it isects the other leg nd so contins the medin. Moreover, the length of the medin is one-hlf the sum of the lengths of the two ses. onversely, the medin of trpezoid is prllel to ech of the two ses nd hs length equl to one-hlf the sum of the length of the ses. ~ iven trpezoid D with D, let line l intersect leg D t midpoint L. Drw digonl. Then l must intersect t midpoint M y the midpoint connector theorem for tringles. pplying this gin to Î, line l must intersect t midpoint N. Segment LN is thus the medin of the trpezoid. strightforwrd rgument estlishes tht L*M*N, so LN = LM + MN. gin y the midpoint connector theorem for tringles, LN = LM + MN = D + = D ( ) L M N D or the converse, let LN e the medin of trpezoid D with L the midpoint of D nd N LN the midpoint of. If is not prllel to the se, construct line r tht goes through L nd is prllel to the se. y the first prt of the theorem, r must contin the medin. Thus the medin is prllel to the ses, nd y the rgument ove hs length equl to one-hlf the sum of the length of the ses. O

3 Theorem (rllel rojection): iven two lines l nd m, locte points nd N on the two lines, we set up correspondence : N etween the points of l nd m y requiring tht ' ', for ll on l. We clim tht this mpping, clled prllel projection, 1) is oneto-one, 2) preserves etweeness, nd 3) preserves rtios of segments on the lines. The fct tht prllel projection is function follows from the rllel ostulte: point cnnot mp to two different points, or there would e two different lines prllel to through point. Tht the mpping is one-to-one lso follows from the rllel ostulte: point N cnnot e mpped from two different points nd, or else there would e two distinct lines through N prllel to. Tht the mpping preserves etweenness follows from the fct tht prllel lines cnnot cross (given **R, if N*RN*N, then is on one side of RR' nd N is on the other; segment ' must cross RR ' ). We spend the rest of our time proving tht the mpping preserves rtios of segments, ' ' specificlly, tht for points nd, =. If the lines l nd m re prllel, then the R ' R ' qudrilterls ecome prllelogrms, nd opposite sides re equl, so = NN nd R = NRN, nd the result is trivil. So we ssume tht the lines meet t point, forming tringle ª. We prove the following Lemm: R ' ' ' R'

4 Lemm: iven tringle ª, if lies on nd on such tht / = /, then. We prove the contrpositive of the sttement. Thus we ssume tht is not prllel to, then show tht / D/D. egin y constructing the prllels nd to through nd, respectively. We hve two cses, ** or **. We will rgue the cse **, the other cse eing nlogous. Since etweenness is preserved under prllel projection, **. Now: isect segments nd, then isect the segments determined y those midpoints, nd so on, continuing the isection process indefinitely. We clim tht t some point, one of our isecting points will fll on, nd the corresponding midpoint will fll on. This is true since for some n, > (the rchimeden roperty of rel numers). So, it is lso true tht for some m, 2 >, so /2 m <.

5 Note tht the segments joining midpoints (like ) re prllel to y the midpoint connector theorems for tringles nd trpezoids. Moreover, the process of repetedly finding midpoints prtitions the segments nd into n congruent segments of length nd, respectively. There re lso some numer k of prllel lines etween nd, (counting ). Then: = k = n = k = n. y lger, k k k k = = nd = = ; therefore, =. owever, ecuse n n n n *** nd ***, < nd <. So, < = <. Thus,. (If it hppens tht **, n exctly nlogous rgument gives us.) > so gin We hve shown the contrpositive of the sttement we wnted to prove. In summry, if =, then. O

6 Now we prove the rest of the rllel rojection Theorem: Theorem (The Side-Splitting Theorem): rllel projection preserves rtios of line segments. Specificlly, if line prllel to the se of ª cuts the other two sides & t points nd, respectively, then / = /, nd / = /. We locte N on such tht N so tht / = N/, nd construct line '. y the preceding theorem, '. ut y hypothesis, so ' =. Therefore, N =, nd / = /. To get the other rtio, note tht ** nd ** so tht = + nd = Then =, or tking reciprocls, =. Thus = 1+, or =. Tking reciprocls gives the desired result. '

7 finl note: verything we hve done here could hve een done on trpezoid insted of tringle. In the cse of trpezoid, the point t the top of the tringle is replced y segment D, nd the mjor theorem is then stted s: iven trpezoid D, if lies on nd on D such tht, then / = D/D. The picture is slightly different, ut the min ides re exctly the sme. D R S ll this forms the foundtion for our study of similrity in the next section.