Monday, August 12, 2013

Problem 913: Right Triangle, Double Angle, Triple Angle, Metric Relations

Geometry Problem
Post your solution in the comment box below

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 913.

Online Geometry Problem 913: Right Triangle, Double Angle, Triple Angle, Metric Relations.

6 comments:

  1. http://img819.imageshack.us/img819/7592/s939.png
    Draw 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

    ReplyDelete
    Replies
    1. 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?

      Delete
    2. Ajit
      Yes,
      By the result of problem 911, BCDE is a cyclic quadrilateral and hence angle CDE = angle CBE= 90
      Peter.

      Delete
    3. My comment was quite unnecessary. I didn't see Problem #911 and went straight to 913. Hence the error.

      Delete
  2. Extend CD to F to form isoceles Tr. ECF with base < 90 - @ and so since < EAF = 2@ EA = AF and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. Mark P on AE such that AP=AC=b
    Extend 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

    ReplyDelete