Sunday, May 18, 2008

Elearn Geometry Problem 6

Geometry Problem, Triangle, Angles

See complete Problem 6 at:

Triangle, cevian, equal segments, and angles. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. CD/BD = sin(6x)/sin(3x)= 2cos(3x)
    AB/BD = sin(3x)/sin(2x)
    Thus, 2cos(3x) = sin(3x)/sin(2x)
    or sin(2x) = (1/2)* sin(3x)/cos(3x)
    Now if x=15 then sin(2x)=1/2 & sin(3x)=cos(3x) which satisfies the equation.
    So x = 15 deg.


  3. bae deok rak( 3, 2010 at 2:10 AM

    Let E be a point with triangles DBA and EDC are congruence(AB=CD,AD=CE,ang(BAD)=x=ang(DCE), ang(ACE)=4x). Then DB=DE and ang(EDC)=2x and hence
    Furthermore, since BA=CD, BE=EC=AD=CA and ang(EBA)=3x=ang(ACD), triangles BAE and CDA are congruence. we see that AE=AD and so triangle AEC is an equilateral. we get 4x=60 degree, that is, x=15 degree.

    1. In case anyone got confused, he swapped A and B. Replace every A with B and vice versa and it should become clear what he meant.
      Also, point E lays below DC in his solution, closer to the D point (at first there are four different orientations for this construction).

  4. Comment on "Joe": Trial-error is not accepted in the Netherlands

    Comment on "bae deaok rak":1)Where did you put point E? 2)AD=CA is not true

  5. Given:
    B=x+90(90 because that is right angle)

    x=15+90 =2x =3x
    =2(15) =3(15)
    =30 =45

    1. El enunciado nunca dice que sea un angulo recto, hay que demostrarlo

  6. To Jonathan (Problem 6): In your statement B=x+90.... 90 is not a data.

  7. if you notice the DBC this is 90%. and why B=x+90 because that is seperated. i'm right??

  8. Video solution (Spanish version) by Eder Contreras and Cristian Baeza at Geometry problem 6. Thanks Eder and Cristian.

  9. The solution is uploaded to the following link:

  10. To Rengaraj PradheepKannah
    In your solution of problem 6, there’s something I didn’t understand.
    The first circle is defined as having center at the midpoint G of AB (with diameter AB) and the second circle has center at the midpoint E of CD (with diameter CD).
    I can’t see why the first circle meets CD in its midpoint E. If we don’t assure that E lies on the first circle, it’s not clear that AG = GE.

  11. To Nilton Lapa
    Greetings to you too.
    Did you notice DBC is an Isosceles triangle.
    Since E is the Mid point of DC BEC is 90 Deg.
    As AB is the diameter of the first circle according to the angle on semi circle is 90 Deg
    The first circle should go thriugh E.


    1. Another way to clear up this confusion about the first circle passing through E:

      ∠EDB is exterior of △ADB, so ∠EDB = ∠ABD + ∠DAB = x + 2x = 3x.
      Therefore △DBC is isosceles (same base angles, ∠EDB = 3x = ∠BCE), that is, BD = BC.
      Drop a perpendicular from B to DC (that is, BE⟂DC). As it is the altitude in an isosceles △DBC, it divides its base into equal pieces, so ED = EC and E is a midpoint of DC.

      Now the part regarding the circle:
      Note that △ABE is a right triangle, since BE⟂DC (and also BE⟂AC, because A,D,C are collinear).
      Since we've chosen G to be the midpoint of its hypotenuse AB (AG = GB), the segment GE is a median in a right triangle △ABE, so it must also be equal to these parts: GA = GB = GE (three radii of the first circle).
      Therefore points A, B, E lie on the same circle centered at G.

      Rengaraj's response assumes that E lies on the circle when it says that it is a right triangle in a circle.

  12. To Rengaraj PradheepKannah
    Thanks. The explanation is absolutly clear. My doubt in problem 6 was enlighted.
    Nilton Lapa

  13. E es el punto sobre D-C tal que AD=DE
    F es tal que B-F paralelo a A-C y BF=AD
    D-B es paralela a E-F
    G es intersección entre E-F y B-C formando EBFC cíclico
    ang(DBC) = 180°-6α
    △FCD es isósceles
    (1) ang(FCD) =ang(DFC)=90°-α
    △BGE es congruente a △FGC por la simetría
    general en los triángulos isósceles △BFE y △EGC
    y el paralelismo de BF con EC.
    ang(CEB) = 90°-α
    Entonces △ABD es semejante a △EFH, por lo tanto
    (2) ang(GEB)=ang(FCG)=2α
    Comparando resutado (1) con (2)
    6α=90°; α=15° Q.E.D.


  14. Let E be a point on the extension AB such that AE=AC.
    Bisect ∠A and let a point F on the angle bisector of ∠A such that m∠AEF=m∠ACF=x.
    Then ∆AFE≅∆AFC by ASA and at the same time they are both isosceles.
    Let a point H on BC such that the angle bisector of ∠A intersect BC at H.
    Then m∠CFH=m∠EFH=m∠FEH=m∠FCG=2x, also m∠EBC=5x and m∠EGC=6x. An exterior angle is equal to the sum of its remote interior angles.
    Again by ASA, ∆EHF≅∆CHF and they are also isosceles triangles, then EH=FH=CH.
    By perpendicular bisector concurrence theorem, CG⊥EF that makes ∠EGC a right angle.
    Therefore: 6x=90°,x=15°

  15. Replies
    1. Couldn't find the G spot, eh? ;)
      Judging by the later statement about "CG⊥EF", I would say it must be the intersection point of EF and CB (just draw it in proportion and knowing the end result, and you'll see).

      Still I don't quite follow the end of the proof, about using the "perpendicular bisector concurrence theorem", since nowhere has been proved that CH and EH are perpendicular bisectors of CF and EF (even if I can see it in the picture). The train of thoughts might be correct, but the proof is still incomplete.

  16. See my proof, maybe similar to the video-one mentioned by the owner of the site on Sept, 4, 2011, but I am not able to see that.

    1. To Stan Fulger
      Spanish solution video at


    I think this is different from the previous solutions.


  19. There are so many solutions, but I think this is a new one
    Drop perpendicular BE on AC ; M on AB with MA=MB=> AM=BM=ME=EC=DE=a
    From ME=AM => ang(AEM)=2x ; from ME=EC => ang(EMC)=ang(ECM)=x => ang(MCB)=2x
    triangle ABC and CBM are similar => BC/MB=AB/BC +> BC^2 = 2a^2 => BC = a sqrt(2)
    In triangle EBC is BE=EC => triangle MEB is equilateral => ang(ABE)=60 => 2x=30 => x=15

    1. It's the same solution as the Turkish one.

  20. angle BDC =angle BCD =3x;
    make E midpoint of DC, BE perpendicular to DC;
    Make K midpoint of AB;
    connect KE;
    so AK=KB=KE =DE=EC;
    because KE=DE=EC' so triangle DKC is right triangle;
    because AK=KE; so angle KEA=angle KAE = 2x;
    so angle ECK =x;
    so DKBC on the same circle,
    so angle DBC =angle DKC =90;
    so 6x =90, x=15;

  21. Replies
    1. In your solution I don't see any proof that the second circle indeed passes through B. Don't depend on GeoGebra's drawings, they can be deceiving; depend on math & logic. Wenlong's solution above contains the extra step needed to prove that DKBC (your DFBC) is cyclic and B indeed lies on the same circle with D, F and C.

  22. B belongs to the first circle (center F)
    <FBD =x intercepts the second circle (center E) on arc FD
    <FEB=2x intercepts the second circle (center E) on arc FD
    This is only possible if B belongs to the second circle
    Therefore B belongs to the two circles

  23. Let E be the mid point of CD and let F be the midpoint of AB

    So AF = FB = FE = DE = CE = c/2

    < AEX = 2x and < EFC = < ECF = x, so < BCF = 2x

    Hence BC^2 = a^2 = BF.AB = c^2/2 = BE^2 + CE^2 = BE^2 + c^2/4
    Therefore BE = c/2 = CE and so < BCE = 3x = 45 and x = 15

    Sumith Peiris
    Sri Lanka

  24. Grab a pen and a piece of paper, and illustrate along.

    Before we begin, let us see what new info we can obtain:
    > Angle ABC = 180° - 2x - 3x = 180° - 5x
    > Angle ADB = 180° - x - 2x = 180° - 3x
    > Angle BDC = 180° - (180° - 3x) = 3x
    > Since angles BDC and BCD are equal, triangle BCD is isosceles.
    > Since triangle BCD is isosceles, line BD = line BC

    Add point E to the figure and connect it to points C and D such that a duplicate of triangle ABD is formed inside triangle BCD. Do note that point E should be closer to point D than point C, or else the solution would not work.

    At this point, we can see that:
    Since triangles ABD and CDE are congruent.
    · Angle CDE = 2x
    · Angle ACE = x
    · Line AD = line DE
    · Line CE = line BC = line BD
    > Angle BDE = 3x - 2x = x
    > Angle ADE = 180° - 3x + x = 180 - 2x

    Now connect point E to point A. After this point we shall use the same solution as we did in problems 4 and 5.

    Notice that:
    > Triangle ADE is isosceles since line AD = line DE.
    > Angle CAE = (180° - (180° - 2x))/2 = 2x/2 = x
    > Triangle ACE is isosceles as angles CAE and ACE are equal.
    > Line AE = line CE because triangle ACE is isosceles.

    Add point F on line AB and connect it to point C such that CF is the angle bisector of angle BCE. Lastly, connect point F to point E.

    We can see:
    > Angle ACF = x + x = 2x = Angle CAF
    > Triangle ACF is isosceles because angles ACF and CAF are equal.
    > Line AF = line CF since triangle ACF is isosceles.
    > Triangles AEF and CEF are congruent by SSS because:
    · Line AE = line CE
    · Line AF = line CF
    · They share side EF.
    > Triangles CEF and CBF are congruent by SAS because:
    · Line CE = line BC
    · Angle ECF = angle BCF = x
    · They share side CF
    > Since triangles AEF, CEF and CBF are congruent:
    · Angles AFE, CFE and CFB are equal.

    Now, we solve the problem:
    > Angle AFE + angle CFE + angle CFB = 3(Angle CFB) = 180°
    > Angle CFB = 180°/3 = 60°
    > Since all 3 angles in a triangle add up to 180°,
    x + 180° - 5x + 60° = 180°
    240° - 4x = 180°
    4x = 60°
    x = 60°/4 = 15°

  25. Considering trinagle BCD

    Considering triangle ABC

    Since CD=AB & BD=BC
    cos4x=1/2 or sinx=0(rej.)