Sunday, May 18, 2008

Elearn Geometry Problem 2

See complete Problem 2
Triangle, Angles, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. En el triangulo ABC,los angulos A=C=40 ; angulo B =100 => lado a=lado b =1.
    En el triangulo AOC ,angulo O=150

    En el triangulo ABC,teorema del seno => b=sen(100)/sen(40)=sen(80)/sen(40)=2cos(40)
    Prolongando la recta AO corta al lado a en Q.En el triangulo AQC,el angulo Q=120 ;en el triangulo OQC; Q=120,C=30 O=30 =>QO=QC
    Como AQ es bisectriz =>QB/AB =QO/AO =>QB/1=(1-QB)/2cos(40) =>QB=(1/(1+2cos(40)) Qc=/(1+2cos(40))

    En el triangulo OQB , angulo Q=60 ; angulo O =120 - X ; aplicando el teorema del seno a OQB ; 2cos(40)/sen(x)=1/sen(120-x) esto es:

    2cos(40)sen(120-x)=sen(x) que trvialmente se vertifica para x=80 (porseno angulo doble..)

  2. prolonga la recta AO de modo que corta al lado BC en el punto I y formarás un triángulo ACD cuyo ángulo ADC es 80º,entonces el triángulo ADC es isósceles, luego prolonga los lados CD y AB hasta que se corten en un punto E,entonces el punto I será el incentro del triángulo AEC.El ángulo AEC es 60º y también el ángulo DIC lo que el cuadrilatero BEDI es cíclico.Se traza el segmento BD y el triángulo BDI es isósceles de 30,120,30,entonces el cuadrilatero BDCO es cíclico por tanto el valor de x es 80º respectivamente.

  3. After translation, I interpret Abrahansom's analysis in English as follows:
    Extend AO to let it intersect BC in point I and further to point D so as to form a triangle ACD whose angle ADC is 80º. Then triangle ADC is isosceles. Now extend AB and CD and let them intersect in point E. Now it is evident that I is the in-centre of triangle AEC. Angle AEC is 60º and also angle DIC = 60º. Quadrilateral BEDI is cyclical. So far so good!
    Join B to D.
    Now triangle BDI is isosceles with angles 30.120.30 (why?)and quadrilateral BDCO is cyclical. Therefore the value of x is 80º.
    Could you please explain as to how the last three conclusions are drawn?

    1. I find the solution very elegant. Thank you to the one who posted on 30 Dec 2008. With respect to the questions of Ajit:

      1. Why is BDI isosceles:
      since BEDI is cyclical, any two angles facing the same arc of the circumscribed circle will have the same values. Therefore <IED=<IBD and <IEB=<IDB. Now using the fact that I is the in-center of AEC we know that IE is the bisector of <AEC. Thus <BEI = < IED = 30 deg
      2. Why is BDCO cyclical:
      Using the same property, any two equal angles passing through the same two points define a cyclic quadrilateral. <OCB =30 deg by definition. In the previous step it was proven that <ODB = 30 deg. Therefore BDCO is a cyclic quadrilateral.
      3. Why is < OBC = 80 deg:
      Using the properties of the circle around BDCO, <OBC = < ODC. The latter is 80deg by construction (ACD is isosceles with two 80 deg angles at the base, remember? :)

  4. En mi opinión no se puede demostrar, pues falta un dato: AO = OB

  5. Dear Anonymous,
    Thank you for your opinion, but the problem doesn't need the condition AO = OB.

  6. I'm Brazilian and me english isn't so good.
    And if u reflect the points A and O over the BC to A' and O' line after that u will have an equilateral triangle BOO', and CO'A congruent to COA what makes OO'A = 150° and by SAS OAO' congruent to AOC and AOO'= 10°, mine problem now is to prove that B belong to OA line and that bokes the problem, if someone have any sugestion i will apreciate,thks

  7. Dear Joe,
    if I interpret Abrahansom's thoughts correctly, his last steps are: BEI=IED=30°, as EI is the angle bisector of E. Then, as BEDI is cyclic, IED=IBD=30°, and DIB=120° follows from DIC=60°. So, the angles in BDI are 30°-30°-120°. Thus, BDO=BCO=30°, and BDCO is cyclic. And from this ODC=x=80°.

  8. my english is so bad sorry :
    1. Extend AO to cut BC at the point E .
    2. AC=6.13 .
    5. IC=6.13*sin20/sin120 => IC=2.42 >> IC=IO=2.42 .
    6. BC=4 , and IC=2.42 >> so BI=4-2.42=1.58 .
    7. AOC=150 ... OC=6.13*sin20/sin150 > OC=4.19 .
    8. BO²=(1.58)²+(2.42)²-(2*1.58*2.42*cos60).
    {Cosine rule} ... so>> BO=√¯4.53 .. BO=2.13 ..
    9. now ... sinX=4.19*sin30/2.13 => sinX=0.985 >>so X=80


  10. Didn't the problem say using "elementary geometry - Euclid's elements"?

  11. To Alex Nguyen: Yes, Euclid's Elements

  12. My solution:

    Mi solución:



  14. To Slice

    I wonder what software do you use to draw geometry figure in your solution ? It look realy nice and neat.
    Peter Tran

  15. to Peter:
    I think you should know what is abaut Geogebra.
    You will discover a whole new world in geometric graphics

  16. That's not Geogebra, it's Corel Draw.

  17. Video solution (Spanish version) by Eder Contreras and Cristian Baeza at Geometry problem 2

  18. Take D on the extension of AB, such that AD = AC. The point E is the intersection of the extension of AO and CD. AE é bissector in triangle isosceles ADC, so angle(AEC) = 90º. In triangle AOC, angle(COE) = 30º, so angle(OCE) = 60º, angle(BCD) = 30º and angle(BDC) = 70º. Triangle OCD is isosceles, because OC = OD and, as angle(OCE) = 60º, triangle OCD is equilateral, so OC = CD. This proves that OBC and DBC are congruent, with angle(DBC) = x. I triangle DBC, we have x + 70º + 30º = 180º and x = 80º.

    1. How is angle BDC=70°

    2. Consider triangle ADC and using angle sum of triangle

  19. The solution is uploaded to the following link:

  20. i have found a new way this is little bit complex and lengthy

    (my english is not soooo good)
    extend AO as the extended portion of AO intersects BC at the point D.draw a perpendicular line BE at the point E on AD.extend BE and it intersects AC at the point H . join O,h and O,C. OC intersects BH at F and DH at G.

    consider two triangle ABE and AHE
    angle BAE= angle EAH = 20degree (given)
    AE is common
    angle BEA=angle AEH=90 degree ( as BH is perpendicular on AD)
    so triangle BAE is congruent with triangle EAH (A.A.S rule)
    so angle ABE = angle AHE .........(1)
    BE=EH .........(2)
    now consider two triangle BED and HED
    ED is common
    angle HED = angle BED = 90 (as BH perpendicular AD)
    BE=EH [from (2)]
    so triangle BED is congruent with triangle HED
    so angle EBD= angle EHD .......(3)
    from (1) and (3) we get
    angle ABE + angle EBD = angle AHE + angle EHD
    or angle ABD = angle AHD
    now angle ABD = 180- angle BAC - angle BCA
    = 180-40-40=100
    so angle AHD=100
    now consider triangle ADH
    angleADH= 180-angle DAH- angle DHA=180-20-100=60
    angle CDH= 180 - angle DAC - angle DCA - angle ADH
    consider two triangle ODG and CDG
    DG is common
    angle ODG = angle CDG = 60
    angle DCG = angle DOG = 30 [angle DOG= angle OAC + angle OCA=20+10=30]
    triangle ODG is congruent with triangle CDG
    angle DGO = angle DGC = 90
    angle OGH = angle CGH = 90
    consider two triangle OGH and CGH
    GH is common
    angle OGH = angle CGH = 90
    OG= CG
    triangle OGH is congruent with triangle CGH
    so angle HOC = angle HCO = 10
    angle EOH = angle EOC + angle COH
    =30+10 = 40
    now consider two triangle BOE and HOE
    OE is common
    BE=EH (previously proved)
    angle BEO = angle HEO = 90 [ as BH is perpemdicular to AD]
    triangle BOE is congruent with triangle HOE
    angle EOH = angle EOB = 40
    angle OBE = 180 - angle BEO - angle EOB
    = 180-90 - 40=50
    angle ABE = 180 - angle BAE - angle BEA
    =180 -90-20 = 70
    angle EBD = 100 - angle ABE
    X= angle OBE + angle EBD
    =50 +30 = 80

  21. Let D be on AC such that OD=DC, then by an angle chase AO=OD.
    Let E be on the opposite side of AB as O such that triangle AOE is equilateral, so
    Also, since angle BAC= angle BCA=40, AB=BC.
    Now angle EAB=40= angle DCB, so triangle EAB and triangle DCB are congruent by SAS.
    Thus EB=BD, so EBOD is a kite. Thus OB bisects angle EOD, so angle EOB=angle BOD=80 and then x=80 by an angle chase.

  22. Take D on (AO so that AD=AC; since tr. ACD is a Langley triangle, it's well known that AO=CD (take an equilateral triangle CDP, P inside tr. ACD, then see that triangles ACO and CAP are congruent, a.s.a. criterion). Take then E on AD, so that CE=CD and tr. AOB and BCE are congruent, s.a.s., giving BE=BO and <OBE=100. Now take G, symmetrical of E about BC, see that ACGO is an isosceles trapezoid; since OC is bisector of <ACG, we get OG=AO=CG and <GEO=30. Since by symmetry BG=BE, B is the circumcenter of tr. GEO, making <GBO=2<GEO=60, i.e. tr BOG equilateral. As G is symmetrical of O about the perpendicular bisector of AC, we infer <GBC=<ABO=(100-60)/2=20, i.e. <OBC=<OBG+<GBC=60+20=80, done.

  23. Drop a perpendicular from C to AO to meet AB in D. Then Tr. DOC is easily seen to be equilateral and thus Tr.s BOC & BDC are congruent SAS and x = < CBD = 80

    Sumith Peiris
    Sri Lanka


  25. It is a special case of (30-t, 30-t, t, 30) where t = 10.,20,10,30

  26. Grab a pen and a piece of paper, and illustrate accordingly.

    Firstly, let us calculate some angles:
    Angle ABC = 180 - 20 - 20 - 10 - 30 = 100
    Angle AOC = 180 - 20 10 = 150

    Extend line AO until it touches line BC. We will name the place they meet as point D.
    So far, we can conclude:
    > Angle ADB = 180 - 20 - 100 = 60.

    Add a point E on AC and connect the point to D, forming line DE, such that a duplicate of triangle ABD is formed. It should have intersected with CO. Now, name the intersection point as P. Connect point O and point E to form line OE. There's a lotta stuff you should know after this point, so brace yourself.
    At this point we can conclude:
    > Angle AED = angle ABC, angle BDA = angle EDA.
    > Triangle COD is an isosceles triangle because angle COD = angle DCO. Line OD = line OC.
    > Angle EDC = 180 - angle BDA - angle EDA = 180 - 60 - 60 = 60
    > Angles ODE and CDE are equal.
    > Triangles OED and DEC are congruent by SAS.
    > Angle DEC = 180 - angle AED = 180 - angle ABD = 180 - 100 = 80
    > Angle DEO = angle DEC = 80 by congruence.
    > Angle OBD = angle DEO = 80 by congruence.

    x = angle OBD = 80.

  27. See the drawing

    < BAC=< BCA=40° => < ABC=100°
    Define D such as DAC is isosceles in A, E in the middle of DC
    DAC is isosceles in A =>AD=AC, < ACD=< ADC,
    O in the angle bisector => OC=OD
    < ACD=180-20-90=70° => < ADC=70° and < BCD=30°
    COD is isosceles in O and < OCD=60° => COD is equilateral => < ODB=60°
    B on the angle bisector of BO=BD and < BOD=< BDO
    < ODB=60° and < ACD=70° => < BDO=10°=< BOD
    => < ABO=20°
    < ABC=< 0BC - < ABO
    Therefore x = 80°