A new circle ?

Pubblicato il 7 Aprile 201726 Aprile 2018 di Alessio Vitali

Let’s consider the triangle $\overset {\triangle}{ABC}$ (here for example we consider an acutangular triangle with AB BC AC but the reader can make changes in all other cases). Let cc’ be the bisector of the angle ACB. On the segment AC we consider AE = AC’. We draw the parallel EF to the base AB meeting CB in F.

claim: C’B=FB: infact AC’:AC=BC’:BC by bisector theorem, AC:AE=BC:FB by Talete’s theorem $\Rightarrow$ AC’:AE=BC’:FB; but AC’:AE=1 by hypothesis $\Rightarrow$ BC’:FB=1 $\Rightarrow$ BC’=FB
claim: the incenter O of $\overset {\triangle}{ABC}$ coincides with the circocenter of $\overset {\triangle}{EFC}$ : infact AO, bisector of the isosceles triangle $\overset {\triangle}{AC'E}$ , is the axis of EC’ and similarly BO is the axis of FC’ and they meet in the circocenter of $\overset {\triangle}{EFC'}$ .
Let’s draw the circle circumscribed to $\overset {\triangle}{EFC'}$ meeting the segments CF, AE, AC’ in P, Q, R.
claim: CP=CE, EP $\bot$ CO: infact, $\widehat{EFC'}= \widehat{FC'B}$ because EF // AB; $\widehat{FC'B}= {90}^{\circ}-\cfrac{\beta}{2}$ considering the rectangular triangle C’UB where U = BO $\cap$ FC’; $\widehat{EPC'}=\widehat{EFC'}$ because both insist on the arc EC’; then $\widehat{EPC'}= {90}^{\circ}-\cfrac{\beta}{2}$ . Similarly $\widehat{C'PF}=\widehat{C'EF}={90}^{\circ}-\cfrac{\alpha}{2}$ . Angle $\widehat{EPF}$ insists on the arc EC’F that is the sum of the arcs EC’ and C’F. Then $\widehat{EPF}$ is the sum of $\widehat{C'EF}$ and $\widehat{C'FE}$ that insist on these arcs. Then $\widehat{EPC}=({90}^{\circ}-\cfrac{\alpha}{2})+({90}^{\circ}-\cfrac{\beta}{2})= {90}^{\circ}-\cfrac{\gamma}{2}$ , because $\alpha + \beta +\gamma = {180}^{\circ}$ . But then the triangle $\overset {\triangle}{CTP}$ , where T=CO $\cap$ EP, has $\widehat{CTP}= {90}^{\circ}$ and then its bisector CO is also its height and so it is isosceles.
claim: EP // QF, QF $\bot$ CO, PR // FC’, PR $\bot$ BO, QR // BC’,QR $\bot$ AO: infact the center O of the circle PEQRC’F is the incenter of ABC and then equidistant from AB, BC, AC. Hence the three cords PF, QE, RC’ are equidistant from the center O and so are all equal. From this the claim follows from an elementary property of circles.
claim: the points of contact of the inscribed circle to $\overset {\triangle}{ABC}$ are the middle points M,N,S, of PF, EQ, RC’: infact O is equidistant from the cords PF, EQ, RC’ and the three distances are OM, ON, OS.
claim: this new circle is homothetic to the inscribed circle: infact they have the same center O, and the ratio of radius is the costant ratio OM:OF=ON:OE=OS:OC’.