tag:blogger.com,1999:blog-6933544261975483399.post7818091677776613163..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 518: Parallelogram, Five Squares, Centers, 45 Degrees, MeasureAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-31358049443631954442010-09-04T16:30:26.703-07:002010-09-04T16:30:26.703-07:00Note that BF=CF=DH=HA and BE=EA=DG=GC
And angle E...Note that BF=CF=DH=HA and BE=EA=DG=GC <br />And angle EBF=angle GCF=angle GDH=angle EAH ( angles equal 90+ angle BCD)<br />Triangles EFB, GCF, GDH and EAH are congruent ( Case SAS)<br />We gave angle BFC=angle EFG= 90<br />So EFGH is a square.<br />Consider quadrilateral HFMD<br />FM=1/2 BM=HD ( ½ of diagonal of congruent squares)<br />And FM//HD ( translation of square center H to square center F)<br />HFMD is a parallelogram<br />EF=HF *SQRT(2)/2 = DM * SQRT(2)/2<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-38064954708719650782010-09-04T13:28:26.176-07:002010-09-04T13:28:26.176-07:001)▲EBF = ▲FCG (SAS) => EF = FG, EFG = 90°
2)FM...1)▲EBF = ▲FCG (SAS) => EF = FG, EFG = 90°<br />2)FMDH paralelogram => MD = FH => EF = (DM√2)/2<br /> ( from ▲EFO, EG meet FH at O)c .t . e. onoreply@blogger.com