User:Cleric K/temp

koko
When two rays from the same homothetic center intersect the circles, each pair of anti-homologous points lie on a circle. Let's consider triangles EQS and EQ'S'. They are similar because both share angle ∠QES=∠Q'ES' and since E is the homothetic center EQ/EQ`=ES/ES' Then ESQ=ES'Q'=alpha. Because of the inscribed angle theorem EP'R'=ES'Q'. R'SQ=180-alpha since it is complementary to ESQ. In QSR'P' Angle QSR'+QP'R'=180-alpha+alpha=180 which means it can be inscribed in a circle. From the Secant theorem follows that EQ*EP'=RS*ER' In the same way it can be shown that PRQ'S' can be inscribed in circle and EP*EQ'=ER*ES'.

The proof is similar for the internal homothetic center I. PIR~P'IR' then RPI=IP'R'=alpha. RS'Q'=alpha (inscribed angle theorem). In PRQ'S' RPQ'=RS'Q'=alpha which means P,R,Q' and S' lie on a circle. Then from intersecting chords theorem RI*IS'=PI*IQ' Similarily QSP'R' can be inscribed in a cricle.