Sunday, May 18, 2008

Elearn Geometry Problem 3

See complete Problem 3
Triangle, Angles, Median, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. Solution to this posted at

  2. In your solution:
    Why "ABC appears to be a right triangle"?
    Why triangles ABD and DBC are congruent?

  3. For lack of better ideas, here's a solution by analytical geometry: Let D be the origin and A:(-1,0) & B;(1,0) and let angles BAC & BCA be 3m & m resply. Also let tan(m)=t. Now BD:y=-x while BC:y=(1-x)t. Thus, B is (t/(t-1),-t/(t-1)). Since co-ordinates of A & B are known we've Slope BA = t/(1-2t) = tan(3m) = (3t-t^3)/(1-3t^2)
    In other words, (1-2t)(3-t^2)= (1-3t^2) which yields three solutions of which the only admissible solutions is t=tan(m)=(2)^(1/2) - 1
    In other words, m=angle BCA = 22.5 deg.

  4. Attempted second explanation at as Antonio pointed out a flaw in the first attempt to prove the value of x.

  5. Are Joe and Ajit synonymous?

  6. They're the same person, not synonymous though!

  7. If you want to use trigonometry then:
    In Tr. ABC -- AC/sin(3x+x)= AB/sin(x)
    or AC/AB =sin(4x)/sin(x)
    In Tr. ABD -- AD/sin(3x+45)= AB/sin(45)
    or AD/AB =AC/2AB =sin(3x+45)/sin(45)
    or AC/AB=2sin(3x+45)/sin(45)
    or sin(4x)/sin(x)=2sin(3x+45)/sin(45)
    And this yields,x=0.3926988845 radians
    = 22.5 degrees as the only admissible solution.


  9. Chose point E on BC so that CAE = x. BEA is exterior angle of AEC so that BEA = 2x and AB = BE. Let F is midpoint of AE clearly triangles AFB and EFB are congruent. If H and G are midpoint of AB and BE respectly then AHF, BHF, BGF and EGF are congruent so that HAF = HBF = GBF = GEF = 2x.
    So finally x + x + 2x + 2x + 2x = 8x = 180 deg  x = 22.5 deg.

  10. Why are AHF, BHF, BGF and EGF congruent?

  11. Thank’s Mr. Antonio for your comment. I will try to justify that.
    Triangle AFB is right at F and HF is median sothat HF = AH = HB. By the same reason we can find that GF = BG = GE.
    AFB and EGF are equilateral triangles so that anggle AFH = anggle EFG = 2x.
    From these facts, HF // BE and AB // BG so that BGFH is a rhombus which has properties HG = BF ……………………………. (1)
    In ABE triangle, because H and G mid points of AB and BE respectly then HG // AE and HG = ½ AE = AF ………………… (2)
    From (1) and (2) we get BF = AF = FE ……………………….. (3)
    From (3) anggle ABF = anggle EBF = 2x.
    We had showed that HB = HF = BG = GF
    From these we get angle HFB = angle BFG so that AHF, BHF, BGF and EGF congruent

    From (3) we can find another solution.
    Because F is midpoint of AE and AF = FE = BF then ABE = 90 deg.
    So that 2x + x + x = 4x = 90 deg  x = 22.5 deg

  12. angle ABD=135-3x & DBC=45-x
    so angle ABD=3z & DBC=z
    area of ABD= area of DBC there fore
    so sin3z/sinz=BC/AB
    due to sine theorem in ABC sin3x/sinx=BC/AB
    there fore sin3x/sinx=sin3z/sinz
    therefore 3-4sin^2(x)=3-4sin^2(z)
    x and z are oblique so x=z
    in triangle ABD 3x+3x+45=135 so x=22.5
    please tell me how to send another picture solutions?(

  13. Correção: de Valdir Marques Faria

    A resolução de Anonymou não é consistente quando justifica a congruência dos triângulos. O ângulo de 45° é determinante na resolução do problema. Assim, teremos:
    No triângulo BAD, triângulo ABD = 135°– x e, no triângulo BDC, triângulo DBC = 45°+x. Aplicando a lei dos senos nos triângulos ABD e BDC, teremos:
    sen(3x)/sen(135°-3x) = sen(x)/sen(45°-x) ...
    sen(3x).sen(45°-x) = sen(135°-3x).senx ...
    sen(3x).(sen45°.cosx – cos45°.senx) = senx(sen135°.cos(3x) – sen(3x).cos135°) ...
    sen(3x).(cosx – senx) = senx.(cos3x + sen3x) ...
    sen(3x).cosx – cos(3x).senx = 2.sen(3x).senx ...
    sen(2x) = cos(2x) – cos(4x) ...
    sen(2x) = cos(2x) + quadr.(sen(2x)) – quadr(cos(2x)) ...
    sen(2x) – cos(2x) = (sen(2x) + cos(2x)).(sen(2x) – cos(2x))
    Se sen(2x) for diferente de cos(2x) teremos:
    sen(2x) + cos(2x) = 1 , e assim x = 0 ou x = 45° o que não convém.
    Se sen(2x) = cos(2x), teremos 2x = 45 e x = 22,5°.

  14. I'm brazilian , and my english i'snt good,but i saw a beautifull solution for it, really fast.However i'll explain in portuguese, please translate it.
    Pronlogando o lado AB, traçamos uma reta partindo de C, quer corte perpendicularmente o prolongamento do lado AB, fprmando o ponto E.Se traçarmos o segmento ED, este será a emediana relativa a hipotenusa do triângulo retângulo ACE, portanto terá a mesma medida de AD.Tendo o ângulo C como 90-3x, aparecerá o triângulo isósceles CDE cujo ângulo EDC também será 90-3x.
    Portanto, 90-3x+90-3x+135=180, logo = -6x=-135, portanto x=22,5

  15. The solution of mr. Anonymous ( is nog correct!
    He says:
    "From these facts, HF // BE and AB // BG so that BGFH is a rhombus which has properties HG = BF ."
    That's NOT true!
    Yes, it's a rhombus, but NOT HG = BF!!!
    It's also cleat that his proof can't be correct, because he doesn't use the fact that angle ADB = 45°. IF his proof was solid, then it would be true voor EVERY triangle with angles x and 3x, that x = 22,5°.

  16. Build quadrilateral ABEC recordable => [DB - bisector [EB - bisector x = 22.5



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


  20. Select Point E at BC so that EAC=x
    So, BEA=2x
    therefore AB=BE (as BEA=BAE=2x)
    Since EAC=ECA=x
    While Drawing a perpendicular to AC from E it meets AC at D
    Now Draw a perpendicular From B to AC,
    Say it is BG
    Since Tri.BGD is 90-45-45
    Now Draw a parallel to DG through E, so it meets BG at F
    so FE=GD
    BEF=ECD=x (Corresponding angles, EF//AC)
    so Tri BAG and Tri.BEF are congruent
    therefore ABG=x
    GBD=45, DBC=45-x
    So, ABC=x+45+45-x
    ABC=90 Deg.
    Now draw a semicircle having its center D and radius DA.
    it passes through A,B,C as AD=DC and ABC=90Deg
    SInce the angle subtended at center =2.angle subtended at circumference
    so, x=22.5 Deg.

    Very nice Question, Thanks Mr.Antonio Gutierrez.

  21. Problem 003 revisited.
    Draw through B a parallel to AC and trough D a perpendicular do AC. These two lines meet at E, and DE meets BC at G.
    If BF is the altitude of ABC, then BEDF is a square.
    We have AG = CG, then ang(GAC) = x and ang(BGA) = ang(BAG) = 2x.
    So ABG is isosceles with AB = BG and triangles ABF and GBE are congruent.
    Since ang(BAF) = 3x and
    ang(BAF) = ang(BGE) = ang (CGD) = 90 – x,
    then 3x = 90 – x, or x = 22,5.

  22. Drop BN ⊥ AC
    cot x - cot 3x = (NC - NA)/BN = 2ND/BN = 2
    sin 3x cos x - cos 3x sin x = 2 sin 3x sin x
    sin 2x = 2 sin 3x sin x = cos 2x - cos 4x
    cos 4x = cos 2x - sin 2x
    cos²2x - sin²2x = cos 2x - sin 2x
    (cos 2x - sin 2x) (cos 2x + sin 2x) = (cos 2x - sin 2x)
    cos 2x + sin 2x = 1 or cos 2x - sin 2x = 0

    If sin 2x = 1 - cos 2x then
    2 sin x cos x = 2 sin²x
    cos x = sin x, x = 45° not admissible

    Hence cos 2x - sin 2x = 0
    tan 2x = 1,
    2x = 45°,
    x = 22.5°

  23. Mi solución con geo clasica :p (en español)

  24. Let ED be the perp bisector of AC, E being on BC. Perp.s from A and E to BD are equal hence Tr ABD is congruent to Tr EBD and the result follows.

    Sumith Peiris
    Sri Lanka

  25. I have a bad English. See Whether you could understand my solution.
    Mark a point E on the side BC such that angle BAE = angle BEA = 2x
    so that BE = BA
    Since Tri AEC is isosceles , the perpendicular drawn from E to BC meet the side BC at D.
    This implies angle BDE = 45 deg.
    Now we shall extend the line DE to F such that DF = AD
    it's obvious that Tri BAD congruent to Tri BDF (S.A.S)
    This implies BF = BA = BE and angle BFD = 3x
    In Tri BEF, angle BEF = angle BFE = 3x
    angle BEF = angle DEC = 3x (vertically opposite angles)
    So, in Tri DEC 3x + x = 90 degree
    4x = 90 degree
    therefore x = 22.5 degree


  27. @NagaratnamKrishnakanth
    Nice solution.

  28. Construct point E, where AE=AC and E lies on BC produced. Hence, triangle ACE is an isosceles triangle where <AEC=<ACE, hence <AEC=x. Let AE//DB.

    Since AE//DB and <DAE is 135° (<DAE+<EAD=180°), <BAF=135°-3x
    On the other hand, by angle sum of triangle AEC, <BAF can also be expressed as 180°-5x.
    Hence, 180°-5x = 135°-3x

  29. 2nd Solution

    Let E be BC such that < EAC = x so that < AEB = 2x = < BAE
    Let F be mid point of AE so that < ABF = < EBF = 90-2x
    But < CBD = 45-x = < DBF = < FDB since FD//BC and hence < ADF = x
    Therefore AF = FD = FE = FB
    Hence AFB is a right isoceles triangle and so 2x = 45 and x = 22.5

    Sumith Peiris
    Sri Lanka

  30. This solution is clearer when illustrated. Grab a pen and a piece of paper.

    Available information:
    > Angle ABD = 45
    > Angle BDC = 135
    > Angle BAC = x
    > Angle ACB = 3x
    > Angle ABC = 180 - 4x

    Firstly, create a duplicate of triangle ABC. Do so such that the new triangle partially overlaps, and shares the same base AC as triangle ABC. You will see that the new triangle has an unmarked point. Name it as point E. Name the intersection of line AE and line BC as point O.

    We can conclude that:
    > Triangle AOC is isosceles.
    > Triangles ABC and ACE have the same angles and lengths.
    > Angle BAE = angle BCE = 3x - x = 2x

    Connect point D to point O, forming line DO. Connect point E to point D, forming line DE, if you have not done so. Connect line B to line E, forming line BE.

    We can see that:
    > Angle AOC = 180 - 2x
    > Angle AOB = angle COE = 180 - (180 - 2x) = 2x
    > Triangle BAO is isosceles because angle BAO = angle AOB = 2x
    > Triangle ECO is isosceles because angle ECO = angle COE = 2x
    > Line AC is parallel to line BE.
    > Triangles AOC and BOE are similar.
    > Triangles AOC and BOE are isosceles.

    Now, create a duplicate of triangle BOE. Do it in such a way that the new triangle shares base BE with triangle BOE, but does not overlap it. There will be one unmarked point. Name that point as point F.
    All illustrations are complete. Prepare for the conclusions afterwards, there are a lot of it.

    Now we solve the problem:

    > Quadrilateral BOEF is a rhombus as line BO = line EO = line BF = line EF.
    > Line AB = line BF = line EF = line CE.

    (i) These show that the obtained figure is half of an equilateral (but not a regular) octagon.

    > Triangles BFE and BOE have the same angles and sides because they are identical.
    > Angle AEF + angle BFE = 180 - 2x + 2x = 180 so lines BF and AE are parallel.
    > Line AB = line EF
    > Therefore, ABFE is an isosceles trapezium.
    > This same test can also be used on quadrilateral BCEF. It is also a trapezium.
    > Since two trapeziums can be formed on the same half, it has 3 or more angles on each half.

    (ii) This shows that the obtained figure is half of an octagon that has 6 or more equal angles.

    > The angle formed when two adjacent points are connected to the centre is always 45.

    (iii) The angle formed when two adjacent points are connected to the centre is always 45.

    Only one octagon fits all three conditions.
    That is the regular octagon -- all sides are equal, and all interior angles are equal to 135.

    180 - 2x = 135
    2x = 45
    x = 22.5

    This may be long-winded and all, but it is the one of the only solutions that do not require one to draw conclusions without proof.

  31. Extend AB to E such that BE=AB.
    Mark F on BC such that m(CAF)=x. Join EF and form the triangle AEF
    From construction, m(EBF)=4x, m(BFA)=2x=> ABF is isosceles and AB=BF
    Since m(EBF)=2m(EAF) and EB=BA=BF => B is the circumcenter of AEF
    =>4x=90 and x=22.5

  32. This comment has been removed by the author.

  33. A very easy approach
    Obviously y=x for the equation to stand
    So 45-x=x