tag:blogger.com,1999:blog-6933544261975483399.post409900922573330943..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 399: Right triangle, Midpoints, Distance, Pythagorean theorem, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-14193573921966707472015-08-26T03:56:59.323-07:002015-08-26T03:56:59.323-07:00Let the parallel thro' D to FG meet EG in H. T...Let the parallel thro' D to FG meet EG in H. Then DH = 2x by applying the mid point theorem to Tr. DEH.<br />Now the diagonals of AECH bisect each other hence the same is a parellelogram and so AH = e and < BAH = 90.<br />So by applying Pythagoras to Tr. ADH 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-37148815311081305822012-03-18T23:25:58.843-07:002012-03-18T23:25:58.843-07:00http://img171.imageshack.us/img171/5231/problem399...http://img171.imageshack.us/img171/5231/problem399.png<br />See attached sketch of above problem <br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-9971471226025262042010-06-26T16:41:40.188-07:002010-06-26T16:41:40.188-07:001. Locate a point H such that CH=d and CH // AB
2....1. Locate a point H such that CH=d and CH // AB<br />2. Note that EH= SQRT (e^2 +d^2) <br />3. Note that Triangle ADG congruent with triangle CHG (case SAS) .<br />4. From step 3 we have Angle AGD= angle CHG and 3 points D, G, H are collinear<br />5. From triangle EDH, FG= ½ EH= ½ SQRT(e^2+d^2)<br /><br />Peter Tran <br />Email: vstran@yahoo.comcltran34https://www.blogger.com/profile/08094245599507131157noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-6227893018046540542009-12-08T19:31:07.085-08:002009-12-08T19:31:07.085-08:00This is a repeat of an earlier problem. Nonetheles...This is a repeat of an earlier problem. Nonetheless, complete the rectangle ABCH and join D to F. Extend DF to meet CH in I. It's evident that DF=FI and hence triangles ADF & CIF are congruent which means that AD=CI=d. Since G & I are midpoints of DE & DI resply., we've 2GF=EI or 2x = √((CI^2+CE^2) = √(d^2+e^2).<br /> I do not recall the earlier problem number but remember that Newzad had posted a similar or the same solution in his blog.<br />Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com