tag:blogger.com,1999:blog-6933544261975483399.post135620973924795193..comments2024-10-13T09:45:37.126-07:00Comments on GoGeometry.com (Problem Solutions): Problem 601: Cyclic Quadrilateral, Incenter, Excenter, RectangleAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-82707550282093094982011-12-18T00:30:03.434-08:002011-12-18T00:30:03.434-08:00can u say how CK=P-ACcan u say how CK=P-ACpranayhttps://www.blogger.com/profile/14678451036203124703noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-18841985828684713322011-05-21T10:57:25.085-07:002011-05-21T10:57:25.085-07:00http://img685.imageshack.us/img685/673/probelm6011...http://img685.imageshack.us/img685/673/probelm6011.png<br /> <br /><br />Draw EQ//BC and MN, GK intersect EQ at P and Q.<br />Since N is the midpoint of LK so P and M are the midpoint of EQ and EG.<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-9961290141915817592011-05-16T04:40:45.408-07:002011-05-16T04:40:45.408-07:00Please elaborate how "M is the midpoint of EG...Please elaborate how "M is the midpoint of EG"Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-23059256388411960602011-05-14T11:25:50.713-07:002011-05-14T11:25:50.713-07:00http://img27.imageshack.us/img27/6927/probelm601.p...http://img27.imageshack.us/img27/6927/probelm601.png<br /><br />A,E,G are collinear and cut arc BC at the midpoint of arc BC.<br />D,F,H are collinear and cut arc BC at the midpoint of arc BC.<br />Let L, N, K are the projection of E, M, G over BC . N is the midpoint of BC.<br />We have BL= p-AC and CK= p-AC where p is half of perimeter of triangle ABC<br />So N is the midpoint of LK and M is the midpoint of EG. <br />Similarly M is also the midpoint of HF.<br />Triangles EBG, ECG, HBF, FCH are right triangles so we have MB=MC=ME=MG=MH=MF<br />Quadrilateral HGFE is concyclically with diameters HF and EG so EFGH is a rectangle.<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com