tag:blogger.com,1999:blog-6933544261975483399.post19982542240282102..comments2024-03-19T00:02:30.728-07:00Comments on Go Geometry (Problem Solutions): Problem 325. Isosceles triangle, Altitude, Incircle, ExcircleAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-47918497719257147042010-03-20T13:04:56.142-07:002010-03-20T13:04:56.142-07:00draw ST tg to F and // to AC
name G, K, L, E meet ...draw ST tg to F and // to AC<br />name G, K, L, E meet AD, AB, BD<br />name M, N, R, P, F meet DC, DS, ST, CT<br />------------------------------------------<br />▲AEG=▲AEK=▲RFT=▲PFT => AK=AG=RT=PT ( anlge & radius )<br />▲GED=▲DEL=▲DMF=▲DNF => GD=DM=DL=DN<br />MC = CP tg from C, NS = SR tg from S<br />------------------------------------------<br />BN = BP => BD + DM = BC + MC =><br />BD + DM = AB + MC (1)<br />------------------------------------------<br />SADB + SBST = SABC + SDCTS<br />(AB+AD+BD)∙r+(BS+ST+BT)∙r=BC∙h+(DC+CT+ST+DS)∙r<br />(AB+AD+BD+BS+BT)∙r=BC∙h+(DC+CT+BT)∙r<br />(AB+AD+BD+BD+DM+NS+BC+CT)∙r=BC∙h+(DC+CT+BT)∙r<br />(3∙AB+AD+BD+MC+NS)∙r=BC∙h+(DC+DS)∙r<br />(3∙AB+AG+DM+BK+DN+MC+NS)∙r=BC∙h+(DC+DS)∙r<br />(4∙AB+DC+DS)∙r=BC∙h+(DC+DS)∙r<br />4∙AB∙r = BC∙h<br /><br />4∙r = h<br />--------------------------------------------c .t . e. onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-78128132634174561822010-03-11T13:13:12.809-08:002010-03-11T13:13:12.809-08:00from a point , F, we can draw one ( only ) perpend...from a point , F, we can draw one ( only ) perpendicular to two parallel lines ( GT//HK )c .t . e. onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-59385482545575753202010-03-02T00:05:57.455-08:002010-03-02T00:05:57.455-08:00Why "T, F, K are at the same line (FK perpend...Why "T, F, K are at the same line (FK perpend to BC, & FT to MT)"???Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-13303338233505388182010-01-18T14:23:48.239-08:002010-01-18T14:23:48.239-08:00let be M midpoint of AC, K point BC meet circle F
...let be M midpoint of AC, K point BC meet circle F<br /><br />draw MT tangent to circle F => <br />T, F, K are at the same line (FK perpend to BC, & FT to MT) => TK diameter ( MT//BK)<br /><br />extend TM to G on AH => MG middle line of tr ACH<br />=> G midpoint of AH =><br /><br />GH = AH = TK = 2r => AH = 4rc .t . e. onoreply@blogger.com