tag:blogger.com,1999:blog-8231784566931768362.post1499570647206680076..comments2023-09-09T08:21:55.454-04:00Comments on MathNotations: Circles, Chords, Tangents, Similar Triangles and that Ubiquitous 3-4-5 TriangleDave Marainhttp://www.blogger.com/profile/13321770881353644307noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-8231784566931768362.post-47294499467452025662008-04-16T22:27:00.000-04:002008-04-16T22:27:00.000-04:00Very nice!KVery nice!<BR/><BR/>KKaz Maslankahttps://www.blogger.com/profile/10215535360917928880noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-67451023380658214392007-11-18T22:27:00.000-05:002007-11-18T22:27:00.000-05:00tc--You are absolutely correct that it is non-triv...tc--<BR/>You are absolutely correct that it is non-trivial to prove what most students would 'assume' is true! From my perspective the key is point F, although you may have found an alternative approach. We do not know that A is equidistant from D and E but we can demonstrate (with some effort) that F is! I'll omit the details for now so that others can play around with it. Note that I did give a hint by mentioning 'altitude on hypotenuse' theorems! Since O is equidistant from D and E it follows that the line containing O and F is the perpendicular bisector of DE! <BR/><BR/>I'm glad you appreciated this subtlety. I think this would prove fairly challenging and you're correct that the rest of the problem is easy to complete. Remember, however, I am suggesting to the student that there is a myriad of approaches here and they need to come up with more than one method in their group. This is one way to stretch the stronger student who believes they have solved the problem in rapid fashion and they just sit there complacently.Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-63810129352666195812007-11-18T20:12:00.000-05:002007-11-18T20:12:00.000-05:00Very nice, Dave. This is pretty cool. As an aside,...Very nice, Dave. This is pretty cool. <BR/><BR/>As an aside, it took me a longer time to prove that DG=GE, and that angle AGD was right, than to do the rest of the problem assuming these were true. <BR/><BR/>TCAnonymousnoreply@blogger.com