Tuesday, January 5, 2010

Problem 414: Triangle, Angles, Altitude, Median, Congruence

Proposed Problem
Click the figure below to see the complete problem 414 about Triangle, Angles, Altitude, Median, Congruence.

Problem 414: Triangle, Angles, Altitude, Median, Congruence.
See also:
Complete Problem 414
Level: High School, SAT Prep, College geometry

17 comments:

  1. http://geometri-problemleri.blogspot.com/2010/01/problem-61-ve-cozumu.html

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  2. mark ang ABE as y and BDC as z

    draw EF perpendicular to AB => ang FEB = z

    FEDB is concyclic, if mark M midpoint of EB =>

    ang FEM = ang EMF = z
    ang MFB = ang FBM = y
    from tr FMB = > z = 2y
    from tr BDC = > z = 60, y = 30

    from tr ABC = > x = 36

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  3. /_BED=90-24=66 deg. and thus /_CBD=/_ABE= 66-x. This makes /_BCA=90-(66-x)=24+x. Triangle ABC gives us: (24+2x)+x+(24+x)=180 or x=33 deg.
    Ajit

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  4. Angle CEB = 66 degrees, see circle with diametr AB - that angle BAC is equal 66/2 = 33 degrees.

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  5. Joe, yuv-k:
    'Angle ABC = 90' is not an original data.

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  6. c.t.e.o:
    If M is midpoint of EB,
    angle FEM = angle EFM

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  7. thanks, I know it now

    but at solution of joe angle B is 2x+24 ???

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  8. Okay I now see my error. I've assumed 66-x = x which has not been proven. Sorry about that.
    Ajit

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  9. in every triangle,at every vertex,the angles (side,atitude),(circumradius,other side) are congruent
    centroid G,circumcenter O lie on the line AM
    AM is the Euler line of triangle ABC
    orthocenter H lies on the Euler line AM and on the altitude AD
    H=A
    BAC is a rigth angle, x=33
    .-.

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  10. To Anonymous:
    Why does circumcenter O lie on the line AM?

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  11. sorry ,i have other point names;
    centroid G, circumcenter O lie on the line BE
    BE is the Euler line of triangle ABC
    orthocenter H lies on the Euler line BE and on the altitude BD
    H=B
    ABC is a rigth angle,x=33
    to prove the first theorem
    forget median BE
    draw the circumcircle of ABC and diameter BF
    ABF is a rigth triangle
    inscribed angles AFB and ACB are congruent
    the triangles AFB and BDC are similar
    angles CBD and OBA are congruent
    that's why in this problem O lies on the line BE

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  12. bjhvash44@sbcglobal.net

    AC is the longest side. A perpendicular @M midpoint passes thru the center of the triangles circumscribed circle so AC is achord of this circle. Only way angle a and a can be equal is when AB is a diameter angle ABC = 90 x=33

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  13. angle(ABE)=66-x, angle(EBD)=90-x.
    angle(BCD)=24+x
    by sine law, in tr(ABE),
    sinx/sin(66-x)=BE/AE,
    and in tr(BEC),
    sin(24+x)/sin(90+x)=sin(24+x)/cosx=BE/EC
    therefore, because AE=EC
    cosx*sinx=sin(66-x)*sin(24+x).
    Answer is x=33

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  14. extend BD to G, BE to H, G and H on circle of ABC
    => arc AH = arc GC => ▲AEH = ▲CEG => GEC = 66° =>
    EC bisector => E center => x = 1/2 BEC = 33°

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  15. In this triangle, altitude and median from B are isogonals.
    This only can happen when the triangle is rectangle in B and solution follows

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  16. it's actually "triangle is right"

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  17. Video solution
    http://youtu.be/3iAHgFDKmUQ

    Greetings go-solvers!

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