Thursday, May 10, 2018

Geometry Problem 1356: Quadrilateral, Triangle, Angle, 30 Degrees, Congruence

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

GGeometry Problem 1356: Quadrilateral, Triangle, Angle, 30 Degrees, Congruence.


  1. Let Angle B=4x, draw angle bisector BE of Angle B such that E lies on CD. Since ABC is isosceles Triangle, BE is also perpendicular bisector of AC. We get Angle BEC= Angle AEB = Angle AED = 60 Deg. Also Angle ABE = Angle CBE = 2x.

    Consider Triangle ABE, ED is external bisector Angle AEB and ADB = Angle AEB/2 = 30 Deg. Hence D must be ex-center of Triangle ABE and BD must be bisector of Angle ABE
    Hence Angle ABD = Angle EBD = x, We get Angle DBC = 3x = 3.Angle ABD .

    1. To Agashe
      In my opinion, the statement of line 4 “Consider Triangle ABE, ED is external bisector Angle AEB and ADB = Angle AEB/2 = 30 Deg.” Is not enough to conclude that D is the ex-center of triangle ABE. Please provide more details . See above for the sketch .

    2. If you draw a circle passing through AB such that the chord AB subtends angle AEB/2 at other points on circle. It will intersect CE at 2 ex-centers of Tr. ABE. According to diagram it must be ex-center opposite to B, hence D must be ex-center of Tr. ABE.

    3. If O is the circumcenter of triangle ACD then triangle OAD is equilateral and O is the reflection of A in BD, thus <DBE=<ABD and BE is perpendicular bisector of AC, done.

    4. To Agashe,

      I still do not understand how you conclude that D is the ex-center of triangle ABE. Can you elaborate more, or post a diagram?


      Stan has an excellent solution. See sketch in the above link for details
      Peter Tran

  2. According to previous coomments and the sketch of Peter
    name P( BE meet circle), H (AP meet BD), L ( BD meet circle)
    => ang ALD = ang AED = 60°, ang LAP = LEP = 30° => AP perpendicular to BD
    from conguence of right triangles BAH and BPH => BH bisector

  3. Bisector angle ABC, intersect with CD at E,
    Reflect triangle ABD along line BE, get triangle BCF; angle CBF =alpha; angle BFC = 30 degree;
    Because angle AEB, angle BEC are 60 degree, so angle AED is 60 degree, so angle CEF is 60 degree;
    So A,E,F on the same line;
    Make equilateral triangle BCG, angle BEC = angle BGC = 60 degree;
    So BCEG on the same circle;
    Because angle GEB = angle AEB=60 degree,
    So G, A,E,F on the same line;
    So angle EGC = angle EBC = half angle ABC;
    Also BG= CG, angle BFC=30 degree = ½ angle BGC;
    So B,C,F on the same semicircle;
    Angle FGC= 2x angle FBC = 2 alpha;
    Angle FGC =angle EGC =half angle ABC =2 alpha;
    So, alpha + theta = 2 x ( 2 alpha);
    So theta= 3 alpha;

  4. Reflect A in BD to create the point X. Then angle ADX is 60 degrees, and triangle ADX is equilateral. Hence angle AXD = 60 degrees.
    Since angle ACD= 30 degrees, it must be on the circumference of the circle centred at X with radius XA. (Angle on circumference is half that at centre)
    Therefore triangles AXB and CXB are congruent (SSS) so angle ABX = angle CBX
    Hence 2alpha = theta - alpha and the required result follows immediately.


    Draw angle bisector BE of ∠ (ABC)
    This bisector will perpendicular to AC and meet AC at E
    So ∠ (CEB)=60=∠ (AEB)
    Draw angle bisector EI of ∠ (BEA) => ∠ (BEI)= ∠ (AEI)= 30= ∠ (ADI)
    Since ∠ (ADI)= ∠ (AEI)= 30 So ADEI is a cyclic quadri.
    Since ∠ (ADI)= ∠ (FEI)=30 => central angle ∠ (AOI)= ∠ (IOF)=60
    Triangle BAO congruent to BFO ( case SAS) => BD is the angle bisector of ∠ (ABE)
    and ∠ (CBD)= 3. ∠ (ABD)

    1. Peter: Can elaborate on this line: (FEI)=30 => central angle ∠ (AOI)= ∠ (IOF)=60 (Yes, ∠(FEI), but I do not see how that implies ∠ (IOF)=60 (F is not on the circle.)

    2. ∠(IOF) is the central angle of ∠(IEF) since I, A, D,E,F is cyclic and O is the center
      but ∠(IEF)=30 so ∠(IOF)=60

      Peter Tran

  6. Given ∠ACD = ∠ADB = 30°, and BA = BC.
    Prove that ∠DBC = 3 times ∠ABD.

    For E on CD, BE bisects ∠ABC. Connect AE. BE intersects AC at F. Then AE = CE. ∠BEC = ∠AEB = ∠AED = 60°. If ∠ABC = 4x, then ∠ABE =∠CBE = 2x. Bisect ∠BEA with EN, which meets BD at I. Connect AI; and because ∠IDA = 30° = 60° / 2 = ∠IEA, then IEDA is cyclic. Then ∠IED = 90° = ∠DAI.

    Produce AI to meet BE at P. Consider quadrilateral PEDA:
    ∠APE = 360° -120° - 90° - ∠EDA = 150° - ∠EDA
    = 60° + ∠DBE.

    Consider Δ ABF: ∠BCF = ∠FAB
    = 90° - 2x; ∠AID = 60°; ∠BIA = 120°; ∠IAB = 60° - ∠ABD.
    ∠FAI = 90° - ∠APF = 30° - ∠DBE. ∠EDB = 60° - ∠DBE.
    ∠PAB = 60° + ∠ABD - 2x = 60° - ∠DBE = ∠IAB = 60° - ∠ABD.
    Thus ∠ABD = ∠DBE = x. And ∠DBC = 3x = ∠ABD * 3.

  7. Take O the circumcenter of triangle ACD; ABCO is a kite, thus BO is angle bisector of <ABC. Triangle ADO is equilateral and O is reflection of A over BD, hence BD is bisector of <ABO, done.

    1. Stan, can you elaborate on "O is reflection of A over BD, hence BD is bisector of <ABO"

  8. See that O, circumcenter of triangle ADC is reflection of A in BD, which solves the problem.

  9. Slight recasting of Stan’s solution with his kind permission

    Let O be the circumcenter of Tr. ACD.
    Then Tr. OAD is equilateral
    ADOB is a kite & < DBO = alpha
    ABCO is also a kite and so theta - alpha = 2.alpha & the result follows

  10. Another Solution

    Let BX be perpendicular to AC, X on AC
    Draw a perpendicular from A to CD to meet BX extended at Y and CD at Z.

    Since CXZY is concyclic, < AYB = 30 & so ABYD is concyclic

    < CAB = 60 & Tr. AXZ is equilateral & since AY = CY, Tr. ACY is also equilateral
    Hence AZ = ZY & so AD = DY.

    Hence in concyclic ABYD, < DBY = alpha and so theta - alpha = 2.alpha which gives theta = 3.alpha

    Sumith Peiris
    Sri Lanka