tag:blogger.com,1999:blog-6933544261975483399.post5450750792539152130..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1124: Triangle, Median, Two Squares, AreaAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-86636857813392914112016-09-13T11:05:41.173-07:002016-09-13T11:05:41.173-07:00Problem 1224
Let area triangle ABC is (ABC).The...Problem 1224<br />Let area triangle ABC is (ABC).Then (BMC)=(ABM) (BM=MC). But <BMA+<DMF=180 so<br />(MDF)/(ABM)=(MF.MD)/(BM.AM)=1.So (BMC)=(MDF) or S_1+(MHPQ)+S_(3 ) =S_2 +(MHPQ)+S_4.<br />Therefore S_1+S_(3 )=S_2+S_4.<br />APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-25601729904350589302015-07-18T22:58:17.201-07:002015-07-18T22:58:17.201-07:00Tr.s MBC & MFD are congruent SAS hence they ar...Tr.s MBC & MFD are congruent SAS hence they are equal in area and the result follows<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-47014379603797248272015-06-19T14:15:06.339-07:002015-06-19T14:15:06.339-07:00Consider rotation anti-clockwise 90° about M,
the...Consider rotation anti-clockwise 90° about M, <br />then D→C and F→B.<br />Thus ΔMDF→ΔMCB and hence congruent. <br /><br />The result S₁+S₃=S₂+S₄ then follows. Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.com