tag:blogger.com,1999:blog-6933544261975483399.post7306013795440421090..comments2022-08-07T08:19:29.767-07:00Comments on Go Geometry: Geometry Problem 1417: Triangle, Cevian, Circumcircle, Incenter, Perpendicular, Angle Bisector, Tangent CirclesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6933544261975483399.post-12650693198571412592019-03-10T11:25:43.523-07:002019-03-10T11:25:43.523-07:00B,I and M are colinear, draw line BIM. We have AM...B,I and M are colinear, draw line BIM. <br />We have AM=CM=MI and MI^2=MF.MT. <br />Hence MI is tangent to circumcircle of tr.IFT. <br />We have Angle MIF = Angle MTI <br />Lets assume Angle ABD=x, then Angle DEF= 90-A/2 - x/2. <br />Also Angle MIF = Angle MTI = Angle BIE = 90 - A/2 -x/2 -[(B/2) - x] = C/2 + x/2 <br />Angle BTI= Angle BTM - Angle MTI = C + B/2 - (C/2 + x/2)=C/2+B/2 -x/2=90 -A/2 -x/2=Angle DEF<br />Since Angle BTI = Angle DEF, quad.BEIT is cylic. <br /><br />We get Angle EBI = Angle ETI = B/2 -x<br />Angle ETF = Angle ETI + Angle FTI = B/2 -x + C/2 +x/2= B/2 + C/2 - x/2= 90 - A/2 - x/2 = Angle DEF = Angle DFE hence DE and DF are tangent to circle Q. <br />Also we have Angle IET= Angle Angle MBT=y <br />Then Angle FTQ=90-y=Angle MTO <br />It means T,Q and O are colinear, hence Circle Q and O touch at point T. <br />Pradyumna Agashehttps://www.blogger.com/profile/10300531209692781145noreply@blogger.com