tag:blogger.com,1999:blog-6933544261975483399.post2407605483450416893..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 166Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-7166611925500565962012-07-11T05:57:48.007-07:002012-07-11T05:57:48.007-07:00To Antonio Ledesma:
Why is it S1+S2+[AFD] = 0,5*[A...To Antonio Ledesma:<br />Why is it S1+S2+[AFD] = 0,5*[ABCD]?<br />(I know that "alfombra" means carpet - "tapete", in portuguese, but I don't know the "alfombra" theorem)<br />Thanks.Nilton Lapanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-18770872967916426842012-07-11T05:49:32.489-07:002012-07-11T05:49:32.489-07:00Solution to problem 166.
Let be h1 and h2 the dist...Solution to problem 166.<br />Let be h1 and h2 the distances from E to lines BC and AD, respectively, and h3 the distance from F to line AD. We have BC = AD.<br />Then S1 = BC.h1/2 = AD.h1/2,<br />S2 = S(ADE) – S(ADF) = AD.h2/2 – AD.h3/2<br />and S = S(ABD) – S(ADF) = AD(h1+h2)/2 – AD.h3/2 = AD(h1+h2-h3)/2.<br />Hence S1 + S2 = AD.h1/2 + AD.h2/2 – AD.h3/2 = AD(h1+h2-h3)/2 = S.Nilton Lapanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-73142889093219758392010-12-12T12:00:16.198-08:002010-12-12T12:00:16.198-08:00Es mucho más fácil: aplicación directa del teorema...Es mucho más fácil: aplicación directa del teorema de la alfombra. <br />Por un lado: [ABD]= S+[AFD]=0'5*[ABCD] <br />Y por otro: S1+S2+[AFD]=0'5*[ABCD]<br />Por tanto, S=S1+S2Antonio Ledesmanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-2060603349759981702009-04-16T22:02:00.000-07:002009-04-16T22:02:00.000-07:00Let A be (0,0), B;(b,h), C:(b+d,h) and D:(d,0) and...Let A be (0,0), B;(b,h), C:(b+d,h) and D:(d,0) and let E be (p,q) Now we can determine the F as <br />(dhp/(ph-qb+qd),dhq/(ph-qb+qd)). Now Tr. AFD =d*dhq/2(ph-qb+qd)= hqd^2/(ph-qb+qd) -----(1)<br />While, Tr. BFE =(1/2)(bdhq/(ph-qb+qd) -bq+dhpq/(ph-qb+qd)-pdh^2/(ph-qb+qd) +ph -pdhq/(ph-qb+qd))and Tr. ECD =(dh+bq-hp)/2<br />Now Tr. BFE + Tr. ECD = hqd^2/(ph-qb+qd) upon addition and simplification. In other words, Tr. AFD = Tr. BFE + Tr. ECD because of (1)<br />In parallelogram ABCD, Tr. ABD = Tr. BDC or S + Tr. AFD = S1 + S2 + Tr. BFE + Tr. ECD or S = S1 + S2 since Tr. AFD = Tr. BFE + Tr. ECD <br />Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-63743265450426877602008-12-24T07:51:00.000-08:002008-12-24T07:51:00.000-08:00draw diagonal AC which meets BD at G and BE at H.a...draw diagonal AC which meets BD at G and BE at H.also join GE.<BR/>proof:we know that in a parallelogram diagonals bisect each other.so AG=GC and BG=GD.<BR/>also median of a triangle divides it into two triangles of equal area.<BR/>[tri BGE]=[tri GED]<BR/>[tri BGH]+[tri GHE]=[tri GEF]+[tri BFD]-----( 1 )<BR/>[tri AEG]= [tri GEC]<BR/>[tri AFG]+[tri GEF]=[tri GHE]+[tri HEC]------(2)<BR/>On adding 1 and 2 equations<BR/>[tri BGH]+[tri AFG]=[tri EFD]+[tri HEC]<BR/>On adding [tri AGB] on either side<BR/>[tri ABF]+[tri BGH]=[tri AGB]+[tri EFD]+[tri BHC]<BR/>but diagonals of a parallelogram divide it into four triangles of equal area.<BR/>[tri AGB]=[tri BGC] substitute in the above equation <BR/>and after simplifications we get the required result. S=S1+S2snighttps://www.blogger.com/profile/11562645714321103640noreply@blogger.com