Saturday, February 21, 2009

Problem 18: Right triangles, Angles, Congruence

Proposed Problem
Problem 18: Right triangles, Angles, Congruence.

See complete Problem 18 at:
gogeometry.com/problem/problem018.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

11 comments:

  1. Let angle theta = t & alpha = a. Since B=90 we've 3a + t = 90 or t = 90-3a. In tr. ADE, DE/AE = sin(2a)/sin(180 -2a-90+3a)=sin(2a)/sin(90+a)=2sin(a)*cos(a)/cos(a)=2sin(a) or DE=2sin(a)*AE. From tr. ABE, BE =sin(a)*AE. Thus, DE=2BE
    Ajit:ajitathle@gmail.com

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  2. It's possible to solve this problem without using trigonometry.
    The key is auxiliary construction:

    Geometry problem solving is one of the most challenging skills for students to learn. When a problem requires auxiliary construction, the difficulty of the problem increases drastically, perhaps because deciding which construction to make is an ill-structured problem. By “construction,” we mean adding geometric figures (points, lines, planes) to a problem figure that wasn’t mentioned as "given."

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    Replies
    1. Given sufficient data any such problem can be solved using Trigonometry. The beauty is to solve using pure geometry alone which I think is the purpose of this website.,

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    2. Given sufficient data any such problem can be solved using Trigonometry. The beauty is to solve using pure geometry alone which I think is the purpose of this website.,

      Delete
  3. If you construct EB extending to the same length ending at F so that EB = BF and then join AF. I am not sure if we can show by triangle similarity that EF = DE which means DE =2 BE

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  5. http://ahmetelmas.files.wordpress.com/2010/05/cozumlu-ornekler.pdf

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  6. http://i1237.photobucket.com/albums/ff480/Evan_Liang/DoubleangleProblem2.jpg
    A much easier way using angle bisector and similarity
    construct a reflection of △ABE △ABKalong the line of reflection AB
    ∴△ABK≅△ABE
    ∴∠KAB=∠BAE=a,AK=AE,KB=BE
    ∵∠EAC=2a=∠KAE
    ∴AE is the angle bisector of ∠KAC
    ∴KE/AK=CE/AC=2BE/AE
    AE/AC=2BE/CE ..................(1)
    ∵∠AED=∠ECA,∠EAC=∠EAC (reflexive prop.)
    ∴△ADE∼△AEC (AA similarity)
    ∴DE/CE=AE/AC ...................(2)
    Use transitive Prop. between (1) and (2)
    ∴DE/CE=2BE/CE
    BE=2BE

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  7. Produce CB to F such that EB = BF
    Join AF
    rtΔABF and rtΔABE congruant (SAS)
    Angle(FAE)=Angle(EAD)=2α
    Angle(AFB)=Angle(AEB)=2α + θ
    Agle(EDC)= 2α + θ =Angle(AFB)
    ADEF concyclic
    DE = EF = 2BE (equal circumference angle equal chord)

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  8. Start like the other solutions and reflect ABE to a new triangle ABF.
    Then add point G on the line AC such that AGE = 90 - a
    tri(EDG) is then isosceles and EG = ED. In addition tri(AEG) is isosclese with AE = AG.
    So tri(AEG) is congruent to tri(AFE) via SAS and FE = EG = ED = 2BE from that congruence.

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  9. Nearly same proof as Anonymous above except I proved AECD con cyclic by noting that AF is tangential to Tr. FDC since AE is tangential to Tr. AEC

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