Answer Happy

Goal Proue SSS Lemma Two cuestioct Circles intersect in Yera, One or two pois If there is exactly one pont of ter seenon, Penthou pont lies on the line jobing the centos of the arcles. . co 0 Thcoren (sas) of the corresponding sicles of the triangle SABC and SAC! have equal feng Thi Then nuo magles one congruent Proof Fast assume AAC and A'C' are not degenerad ( means A,B,Core not callinear) There exists on sametty + Such that PCA). A f(0) Band (c) = C' ade NTS 7 By axiom 6. There exist on some my f Buck mat. & CA). A'. Since on sometry and The Holla'e'l, u have JA . 1.CA) f) LAI Ae A By curicam 7 fr67 There_exists on Isometry to such that 4,CA). A Land (£)) B. c 4.00)) - we're done On the other hand, If 4,0) + ' consider the concle centered at A'wuh indius LACI anaithe arce centered at B! wch.cadius LBCI A B Samstore But falfa ces BV TO BE
The candles intersea of most 2 pots by the femma Onect which is cand the cher which must be tch(c)) By axiom 8 feners exests ane sometry to which free every point on AB bu fices nachher pons h(f.coll 6 Sinces fois an iconic hy and since ch is not the point must must mapped officc.) and vice Versa Comparing these tficee samarres we have f (426F.CAN = A' tste fotol) - B! 4. (f,(c))) . c' 7 A ABC = A'B'C' Exercise Degenerate case of sss? 413 SAS 414_ASA

The circles intersect in af most 2 points by the femma one of which is c' and the other which must be f (h(c). By axi.com 8 there is exists are and somety fg ahvon fines every part on Alex but fices.nother pons Joh (f. (c)) Ida He Sinces of is an romony and since Clis not A'B' the point ! Must pist mapped to fcc)) and vice Versa Composing these tfree saman res, we have fo(HCA) - A! toffet foto) B! * Clic - c! A ABC 5 AAC' Exercise De generate case of SSS ? 113 SAS 114 ASA