excenters and excircles

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1 21 onurrene IIi

2 2 lesson 21 exenters nd exirles In the first lesson on onurrene, we sw tht the isetors of the interior ngles of tringle onur t the inenter. If you did the exerise in the lst lesson deling with the orthi tringle then you my hve notied something else tht the sides of the originl tringle re the isetors of the exterior ngles of the orthi tringle. I wnt to led off this lst lesson on onurrene with nother result tht onnets interior nd exterior ngle isetors. THM: exenters The exterior ngle isetors t two verties of tringle nd the interior ngle isetor t the third vertex of tht tringle interset t one point. F F F

3 onurrene IIi 3 F F F F F roof. let nd e the lines iseting the exterior ngles t verties nd of. They must interset. lel the point of intersetion s. now we need to show tht the interior ngle isetor t must lso ross through, ut we re going to hve to lel few more points to get there. let F, F, nd F e the feet of the perpendiulrs through to eh of the sides,, nd, respetively. Then, y s, F F F F. Therefore F F F. Here you my notie prllel with the previous disussion of the inenter, like the inenter, is equidistnt from the lines ontining the three sides of the tringle. y H l right tringle ongruene, F F. In prtiulr, F F nd so is on the isetor of ngle. There re three suh points of onurrene. They re lled the exenters of the tringle. sine eh is equidistnt from the three lines ontining the sides of the tringle, eh is the enter of irle tngent to those three lines. Those irles re lled the exirles of the tringle.

4 4 lesson 21 ev s Theorem y now, you should hve seen enough onurrene theorems nd enough of their proofs to hve some sense of how they work. Most of them ultimtely turn on few hidden tringles tht re ongruent or similr. Tke, for exmple, the onurrene of the medins. The proof of tht onurrene required 2 : 1 rtio of tringles. Wht out other triples of segments tht onnet the verties of tringle to their respetive opposite sides? Wht we need is omputtion tht will disriminte etween triples of segments tht do onur nd triples of segments tht do not. let s experiment. Here is tringle with sides of length four, five, nd six. = 4 = 5 = 6. s n esy initil se, let s sy tht one of the three segments, sy, s n esy initil se, let s sy tht one of the three segments, sy, is medin (in other words, tht is the midpoint of ). now work kwrds. sy tht the triple of segments in question re onurrent. Tht onurrene ould hppen nywhere long, so I hve hosen five points i to serve s our smple points of onurrene. one those points of onurrene hve een hosen, tht determines the other two segments one psses through nd i, the other through nd i. I m interested in where those segments ut the sides of. lel: i : the intersetion of i nd i : the intersetion of i nd

5 onurrene IIi Here re the mesurements (two deiml ple ury): i : i i i i out of ll of tht it my e diffiult to see useful pttern, ut ompre the rtios of the sides i / i nd i / i (fter ll, similrity is ll out rtios). i : i / i i / i They re the sme! let s not jump the gun though wht if isn t medin? For instne, let s reposition so tht it is distne of one from nd three from

6 6 lesson 21 i : i i i i i / i i / i The rtios re not the sme. look refully, though the rtios i / i re lwys three times the orresponding rtios i / i (other thn it of round-off error). Interestingly, tht is the sme s the rtio /. let s do one more exmple, with = 1.5 nd = i : i : i i i i i / i i / i

7 onurrene IIi 7 one gin, the rtios i / i ll hover out 1.67, right t the rtio /. Wht we hve stumled ross is lled ev s Theorem, ut it is typilly given it more symmetril presenttion. ev s THeoreM Three segments,, nd, tht onnet the verties of to their respetive opposite sides, re onurrent if nd only if = 1. roof. similr tringles nhor this proof. To get to those similr roof. = similr tringles nhor this proof. To get to those similr tringles, though, we need to extend the illustrtion it. ssume tht,, nd onur t point. Drw out the line whih psses through nd is prllel to ; then extend nd so tht they interset this line. Mrk those intersetion points s nd respetively. We need to look t four pirs of similr tringles.

8 8 lesson 21 They re: 1. = 2. = = 4. = = 3. = lug the seond eqution into the first nd the fourth into the third = = set these two equtions equl nd simplify = = = 1.

9 onurrene IIi 9 = similr tti works for the other diretion. For this prt, we re going to ssume the eqution = 1, nd show tht,, nd re onurrent. lel : the intersetion of nd Q: the intersetion of nd. In order for ll three segments to onur, nd Q will tully hve to e the sme point. We n show tht they re y omputing the rtios / nd Q / Q nd seeing tht they re equl. Tht will men tht nd Q hve to e the sme distne down the segment from, nd thus gurntee tht they re the sme. gin with the similr tringles: 1. = 2. = = Q 4. = = Q 3. Q Q Q Q =

10 10 lesson 21 lug the seond eqution into the first = nd the fourth eqution into the third now divide nd simplify Q Q = Q Q = = = 1. Therefore / = Q / Q, so = Q. ev s Theorem is gret for onurrenes inside the tringle, ut we hve seen tht onurrenes n hppen outside the tringle s well (suh s the orthoenter of n otuse tringle). Will this lultion still tell us out those onurrenes? Well, not quite. If the three lines onur, then the lultion will still e one, ut now the lultion n misled it is possile to lulte one when the lines do not onur. If you look k t the proof, you n see the prolem. If nd Q re on the opposite side of, then the rtios / nd Q / Q ould e the sme even though = Q. There is wy to repir this, though. The key is signed distne. We ssign to eh of the three lines ontining side of the tringle diretion (sying this wy is positive, this wy is negtive). For two points nd on one of those lines, the signed distne is defined s if the ry points in the positive diretion []= if the ry points in the negtive diretion. 1 0 Signed distne from. The sign is determined y hoie of diretion

11 onurrene IIi 11 This simple modifition is ll tht is needed to extend ev s Theorem ev s THeoreM (extended VersIon) Three lines,, nd, tht onnet the verties of to the lines ontining their respetive opposite sides, re onurrent if nd only if [] [] [] [] [] [] = 1. Menelus s Theorem ev s Theorem is one of pir the other hlf is its projetive dul, Menelus s Theorem. We re not going to look t projetive geometry in this ook, ut one of its key underlying onepts is tht t the level of inidene, there is dulity etween points nd lines. For some very fundmentl results, this dulity llows the roles of the two to e interhnged. Menelus s THeoreM For tringle, nd points on, on, nd on,,, nd re oliner if nd only if [] [] [] [] [] [] =

12 12 lesson 21 roof. = suppose tht,, nd ll lie long line. The requirement tht,, nd ll e distint prohiits ny of the three intersetions from ourring t vertex. ording to sh s lemm, then, will interset two sides of the tringle, or it will miss ll three sides entirely. either wy, it hs to miss one of the sides. let s sy tht missed side is. There re two wys this n hppen: 1. intersets line on the opposite side of from 2. intersets line on the opposite side of from The two senrios will ply out very similrly, so let s just look t the seond one. Drw the line through prllel to. lel its intersetion with s. Tht sets up some useful prllel projetions. From to : From to :. ompring rtios, =. ompring rtios, = nd so nd so =. =.

13 onurrene IIi 13 Just divide the seond y the first to get 1 = = =. Tht s lose, ut we re fter n eqution tht lls for signed distne. so orient the three lines of the tringle so tht,, nd ll point in the positive diretion (ny other orienttion will flip pirs of signs tht will nel eh other out). With this orienttion, if intersets two sides of the tringle, then ll the signed distnes involved re positive exept [] =. If misses ll three sides of the tringle, then three of the signed distnes re positive, ut three re not: []= []= []= either wy, n odd numer of signs re hnged, so [] [] [] [] [] [] = 1. = let s turn the rgument round to prove the onverse. suppose tht [] [] [] [] [] [] = 1.

14 14 lesson 21 Q Drw the line from tht is prllel to nd lel its intersetion with s. There is prllel projetion from to so tht nd therefore =. Drw the line from tht is prllel to, nd lel its intersetion with s Q. There is prllel projetion from to so tht Q nd therefore Q = now solve those equtions for nd Q, nd divide to get [Q] [] [] [] = [] [] [] [] = [] [] [] [] [] [] = ( 1)=1. oth nd Q re the sme distne from long. Tht mens they must e the sme.

15 onurrene IIi 15 The ngel point k to exirles for one more onurrene, nd this time we will use ev s Theorem to prove it. THe ngel oint If,, nd re the three exirles of tringle so tht is in the interior of, is in the interior of, nd is in the interior of ; nd if F is the intersetion of with, F is the intersetion of with, nd F is the intersetion of with ; then the three segments F, F, nd F re onurrent. This point of onurrene is lled the ngel point. roof. This is tully pretty esy thnks to ev s Theorem. The key is similr tringles. lel, the enter of exirle,, the enter of exirle, nd, the enter of exirles,. y tringle similrity, F F F F F F. F F F

16 16 lesson 21 F F F ev s Theorem promises onurrene if we n show tht F F F F F F = 1. Those tringle similrities give some useful rtios to tht end: F F = F F F F = F F F F = F F. so F F F F F F = F F F F F F = F F F F F F = 1. y ev s Theorem, the three segments re onurrent.

17 onurrene IIi 17 exerises 1. use ev s Theorem to prove tht the medins of tringle re onurrent. 2. use ev s Theorem to prove tht the orthoenters of tringle re onurrent. 3. Give ompss nd stright-edge onstrution of the three exirles nd the nine-point irle of given tringle. If your onstrution is urte enough, you should notie tht the exirles re ll tngent to the nine-point irle ( result ommonly lled Feuerh s Theorem).

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