Geometry Problem
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Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 913.
Monday, August 12, 2013
Problem 913: Right Triangle, Double Angle, Triple Angle, Metric Relations
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http://img819.imageshack.us/img819/7592/s939.png
ReplyDeleteDraw F on CD such that CD=DE=e
Note that ED perpendicular to CF and triangle CEF is isosceles
In cyclic quadrilateral BCDE , angle(CED)= angle(CBD)= alpha= angle(DEF)
Triangle EAF similar to triangle CEF ………(case AA)
So triangle EAF is isosceles => d= B+2d
Peter, don't you need to prove that ED is perpendicular to CD? It's easy to do so but still it needs to be demonstrated, does it not?
DeleteAjit
DeleteYes,
By the result of problem 911, BCDE is a cyclic quadrilateral and hence angle CDE = angle CBE= 90
Peter.
My comment was quite unnecessary. I didn't see Problem #911 and went straight to 913. Hence the error.
DeleteExtend CD to F to form isoceles Tr. ECF with base < 90 - @ and so since < EAF = 2@ EA = AF and the result follows
ReplyDeleteSumith Peiris
Moratuwa
Sri Lanka
Mark P on AE such that AP=AC=b
ReplyDeleteExtend EC to Q such that m(CPQ)=@ (read as Alpha) and hence Tr.PQE is similar to Tr.CBD (as CBED is cyclic)
Let PQ meet AC at R. Observe m(RCB)=m(RPB)=90-2@ hence RCPB is cyclic and m(CRP)=m(CBP)=90
since m(CPQ)=m(PQC)=@ and PR _|_ to AC=> PR is mid-point of PQ => PQ=2.PR
Per construction AP=b hence Tr.APR is congruent to Tr.ACB => CB=PR
As it is alreay established that Tr.PQE is similar to Tr.CBD and since CB=PR=PQ/2=>PE=2.CD=2e
Hence AE=d=AP+PE=b+2e
Simple solution:
Extend CD to F such that DF=e. since CD _|_ CF (as CBED is cyclic) and CD=DF=e=> CEF is isosceles and m(EFD)=90-@=>m(AEF)=90-@
Hence AEF is isosceles => d=b+2e