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Re: similar triangles
Posted:
Aug 1, 1996 5:46 PM


Tim Doskoch wrote: > > We know that if two angles of one triangle are equal to two angles of > another triangle, then the triangles are similar, and corresponding sides > have the same ratio. > > So if we have two triangles ABC and DEF (labeled clockwise from the apex), > where angle A = angle D and angle B = angle E then: AB/DE = BC/EF = CA/FD. > > Often, somebody will use AB/BC = DE/EF and get the same answer as if they > used the above ratios. > > In what circumstances can you mix up the similar ratios with a correct result? > > This also begs the question, in what circumstances will you not get a > correct result? > > Tim Doskoch > Edmonton, Alberta > Canada
Since we may assume BC and EF <> 0, I guess I see the two statements:
AB/DE = BC/EF and AB/BC = DE/EF
as equivalent!
AB/DE = BC/EF > AB*EF = DE*BC > AB*EF/BC=DE > AB/BC = DE/EF
 Howard L. Hansen Assistant Professor of Mathematics Western Illinois University Macomb, IL 61455 http://www.ECNet.Net/users/mfhlh/wiu/index.htm "Good mathematics is not how many answers you know, but how you behave when you don't know the answer."



