Tuesday, February 12, 2013

Problem 857: Triangle, Median, Midpoint, Angles, 30, 45 Degrees, Congruence, Auxiliary Lines

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

Click the figure below to see the complete problem 857.

Online Geometry Problem 857: Triangle, Median, Midpoint, Angles, 30, 45 Degrees, Congruence, Auxiliary Lines

Posted by Antonio Gutierrez at 11:43 AM
Labels: , , , , , ,

5 comments:

  1. Michael TsourakakisFebruary 12, 2013 at 4:55 PM

    CA ,perpendicular to AB. Then, angle ACE=60 ,and angle ECB=45.
    Because, angle EBC=45 ,EC=EB=ED=DA. So , 2x=30 ,x=15 degrees
    http://img571.imageshack.us/img571/3688/p857trianglemedianmidpo.gif

    ReplyDelete
  2. W FungFebruary 12, 2013 at 5:00 PM

    Reflect C along AB to a point E.

    Angle(EBC)=2*Angle(ABC)=2*(45)=90

    By reflection, Triangle(AEC) is equiliteral and
    since AD=DC, Angle(EDC)=90.

    So, E,D,B,C are concyclic.

    x = Angle(ABC)-Angle(DBC)
    = 45 - Angle(DEC)
    = 45 - (1/2)*(Angle(AEC))
    = 45-30
    = 15.

    ReplyDelete
  3. Peter TranFebruary 12, 2013 at 5:30 PM

    http://img42.imageshack.us/img42/6031/problem857.png

    Draw CE ⊥AB
    We have ∠ (ACE)= 60 and ED is a median of right triangle AEC
    So ED=DA=DC
    We have ∠ (ECB)=105-60=45 => EC=EB=ED
    ∠ (AED)=30=2.x =>x=15

    ReplyDelete
  4. EditorApril 15, 2013 at 7:42 AM

    This kind of problems usually are solvable by placing the circumcenter.
    - Let O be the circumcenter of (ABC).
    - AD=OD=CD, so tr DOB is isosceles.
    - tr OBC is equilateral.
    - ODCB is a deltoid, so BD is perp to OC and <OBD=30°.
    - <OBA=<OAB=15°.
    Finally <x=15°.

    ReplyDelete
  5. Μιχάλης ΝάννοςOctober 10, 2013 at 7:04 AM

    http://www.mathematica.gr/forum/viewtopic.php?f=20&t=35316&p=162751

    ReplyDelete

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