tag:blogger.com,1999:blog-6933544261975483399.post5204235027971408621..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1182: Two Squares, Four Quadrilaterals, Sum of the Areas, Metric RelationsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-62882300631631961982016-01-29T13:40:47.775-08:002016-01-29T13:40:47.775-08:00http://s21.postimg.org/nszfowro7/pro_1182.png
Dra...http://s21.postimg.org/nszfowro7/pro_1182.png<br /><br />Draw new square E’F’G’H’ as shown on the sketch<br />Let x= FF’=GG’=HH’=EE’<br />Define u, v, s, t as shown <br />Denote S(XYZ)= area of XYZ <br />Let S5= S(GFF’)=S(FEE’)=S(EHH’)=S(HGG’)<br />1. Note that u+v=s+t<br />With algebra calculation we get <br /> .a^2 +c^2= u^2+v^2+s^2+t^2+2.x^2+2x(s+t)<br />And b^2+d^2= u^2+v^2+s^2+t^2+2.x^2+2x(u+v)<br />Since u+v= s+t => a^2 +c^2= b^2+d^2<br />2. Let S1’=S(BF’E’A), S2’=S(BF’G’C), S3’=S(CG’H’D) , S4’=S(AE’H’D)<br />Since u+v=s+t => S1’+S3’=S2’+S4’…. (1)<br />We have S1=S1’+S5+1/2x(t-s)<br />And S3=S3’+S5+1/2x(s-t)<br />So S1+S3=S1’+S3’+2.S5…. (2)<br />Similarly S2+S4=S2’+S4’+2.S5…(3)<br />Compare (1),(2) and (3) we will get S1+S3=S2+S4<br /><br /> <br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-10196826093865192622016-01-28T13:38:02.762-08:002016-01-28T13:38:02.762-08:00draw from E parallel to AD and to AB, at the same ...draw from E parallel to AD and to AB, at the same way from F, G, H<br />Sideways EF, FG,GH,HE are formed 4 congr triangles, so and sideways<br />of a, b, c, d <br />And results is clearc.t.e.onoreply@blogger.com