Sunday, May 18, 2008

Elearn Geometry Problem 5

Geometry Problem, Triangle, Angles

See complete Problem 5 at:
www.gogeometry.com/problem/problem005.htm

Triangle, Angles, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

9 comments:

  1. Triangle CDB is isosceles since BD=DC, therefore angle BCD=angle DBC=2x.

    Let point E be a point inside ABD triangle so that AED triangle is congruent to CDB triangle, where AE=ED=BD=DC and as it's given AD=BC. Note that both angles EAB and EAD are equal 2x.
    Angle ADB = 4x=>angle EDB=2x. Triangle EDB is isosceles since ED=BD.
    Angle BAE = angle BAD - angle EAD = 3x - 2x = x.

    Let's reflect point E over AB calling it F. Triangles AFB and AEB are congruent. AF=AE, FB=EB. Angle FAE = 2 * angle BAE = 2x.

    FAE and BDE are congruent isosceles triangles since AF=AE=DE=DB and angle FAE=angle EDB=2x
    =>FE=EB, but FB=EB => triangle FBE is equilateral => angle FBE=60=>AEB=30.

    Angle EBD = (180 - angle EDB) / 2 = 90 - x.
    Angle ABC = 180 - angle BAC - angle BCA = 180 - 5x.
    But angle ABC = angle ABE + angle EBD + angle DBC, or
    180 - 5x = 30 + (90 - x) + 2x=> 180 - 5x = 120 +x => x=10

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    Replies
    1. "Note that both angles EAB and EAD are equal 2x"

      What? How come? If you subtract <EAD=2x from <BAD=3x, the remainder <BAE=x, not 2x.

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

    ReplyDelete
  3. Video solution (Spanish version) by Eder Contreras and Cristian Baeza at Geometry problem 5. Thanks Eder and Cristian.

    ReplyDelete
  4. Draw an angular bisector of BDA.
    Draw DAE=2x, so that the angular bisector of BDA meets it at E.
    So, AE=ED (EAD=EDA=2x)
    But ∆EAD and ∆BDC are congruent under A.A.S case)
    So, ED=BD
    Now, Draw a Circle having the center D and Radius DC
    Obviously it passes through E, B and C
    Let AC and the circle meet at G
    So, GD=BD (Radius)
    Now connect BG
    H be the point of intersection of BG and DE
    Since ∆BDG is an Isosceles DH is perpendicular to BG
    So, HGD=90-2x
    Let F be the reflex of E over AB
    As AF=AE, FAB=EAB=x, AB=AB
    ∆AFB and ∆ AEB are congruent
    So FB=EB
    Also, ∆AFE and ∆BDE are congruent (BD=AF, BDE=FAE=2x, DE=AE)
    So, BE=FE
    Eventually EB=EF=BF
    So, ∆BFE is an equilateral
    Therefore FBE=600
    Since FE and AB are perpendicular to each other
    ABE=300
    Since, Ang.ABC=ABE+EBD+DBC,
    ABC=300+900-x+2x
    ABC=1200+x
    But GBC=90Deg as it is an angle on semicircle (Note that GC is a diameter)
    So, ABG=120+x-90
    ABG=30+x
    Since BGD=BAG+ABG
    900-2x=3x+300+x
    600=6x
    So, x=100
    Note: Since ABD = 120-x the proof for the “PROBLEM 4 is inbuilt”

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  5. Extend BD to E such that BD = DE
    Verify BCE is a right angle; Angle BDA = 4x; ABD = Pi - 7x
    Now cos 2x = BC / BE = BC / 2BD
    => 2 cos 2x = BC / BD = AD / BD = sin (pi - 7x) / sin3x = sin 7x / sin 3x
    => 2 sin 3x cos 2x = sin 7x
    => sin 5x + sin x = sin 7x
    => sin 7x - sin 5x = sin x
    => 2 cos 6x sin x = sin x
    => cos 6x = 1/2
    => 6x = pi/3
    => x = pi/18 = 10 deg

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  6. Dear Antonio - is there a proof using properties of 80-20-80 triangles?

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  7. https://www.youtube.com/watch?v=MeEZakv8WxI

    ReplyDelete