tag:blogger.com,1999:blog-6933544261975483399.post4368443318697209926..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 946: Triangle, Quadrilateral, Area, Diagonal, Midpoint, ParallelAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-83478820878584197722017-05-25T07:48:35.948-07:002017-05-25T07:48:35.948-07:00GH//BD//EF
Further ∆BGF has sides = to half the ...GH//BD//EF<br /> <br />Further ∆BGF has sides = to half the sides of ∆ABD.<br /> <br />So S(EHF) = S(EGF) = ¼ S(ABD)<br />Moreover S(ECF) = ¼ S(BCD).<br /> <br />Hence S(CEFH) = S(EHF) + S(ECF) = ¼ {S(ABD) + S(BCD)} = ¼ S(ABCD) =7<br /> <br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-48725061569537045552013-12-19T17:28:25.365-08:002013-12-19T17:28:25.365-08:00Let S(ABC) denote the area of ABC, and so on.
Si...Let S(ABC) denote the area of ABC, and so on. <br /><br />Since EF//BD//GH, so S(CEHF) = S(CEGF). <br /><br />Consider quadrilateral CEGF. <br />CE = 1/2 CB<br />EG = 1/2 BA<br />GF = 1/2 DA<br />FC = 1/2 DC<br />So CEGF~CBAD. <br /><br />Hence, S(CEHF) = S(CEGF) = 1/4 S(CBAD) = 7Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-48086159356722493102013-12-19T10:46:11.173-08:002013-12-19T10:46:11.173-08:00let's take the informal pragmatic approach: if...let's take the informal pragmatic approach: if the problem statement does not depend on A, B, C, D position but only on quadrilateral area, it holds for a particular case where, say, B=C=E. In this degenerate configuration, it's easy to see that area(CEFH) = area(ABCD)/4 = 7. Hence this must hold for *any* ABCD ;-)<br /><br />bleaugAnonymousnoreply@blogger.com