tag:blogger.com,1999:blog-6933544261975483399.post5282506689794767304..comments2024-04-22T04:55:16.794-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 937: Triangle, Circumcircle, Circles, Tangent, Sides, MidpointAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-79700353353546996732013-12-02T23:08:23.264-08:002013-12-02T23:08:23.264-08:00Apply inversion on B in a way such that line AC is...Apply inversion on B in a way such that line AC is interchanged with circum circle O <br /><br />Line BA and line BC are unchanged whilst circle Oa and circle Oc becomes straight line through A and C , parallel to BC and BA respectively. <br /><br />So the inversion image forms parallelogram B-A-C-D, as D is the intersection of the two inverted lines.<br /><br />Note that point E is inverted along BDE and now it is lying on AC, which is the intersection of diagonal of the parallelogram.<br /><br />The Remaining of proof is trivial by inversion property .<br /><br />QEDnakahttps://www.blogger.com/profile/11277356476170372732noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-27109706181895520722013-12-02T13:28:07.016-08:002013-12-02T13:28:07.016-08:00OOc is perpendicular to AB and OOa is perpendicula...OOc is perpendicular to AB and OOa is perpendicular to BC so <OcOOa=180-<B. OcBOa=OcDOa=180-<B as well, so OcOaDO is cyclic. BOaOOc is parallelogram so OOc and DOa are both radii of circle Oa. That makes OcOaDO an isosceles trapezoid with DO parallel to OcOa both perpendicular to BD. Therefore D must be midpoint of chord BE.Ivan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-44386245100611525972013-12-02T13:03:30.439-08:002013-12-02T13:03:30.439-08:00http://img41.imageshack.us/img41/4263/7fbb.png
Dr...http://img41.imageshack.us/img41/4263/7fbb.png<br /><br />Draw points M, L and F per attached sketch.<br />We have right angles BCL, BCF, BAM, BAF, CBM and ABL.<br />So M, A and F are collinear . L, C and F are collinear<br />And BM// FL and BL//MF => MBLF is a parallelogram and diagonals BF and ML cut at midpoint O of BF<br />Since OaOc is a perpendicular bisector of BD => M, D and L are collinear and OD perpen. To BE<br />So D is the midpoint of BE <br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com