Thursday, April 5, 2012

Problem 739: Triangle, Double Angle, External Angle Bisector

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 739 details.

Online Geometry Problem 739: Triangle, Double Angle, External Angle Bisector.



    Extend BC to E such that BE=BA ( see sketch)
    Note that ∠ (ABD)= ∠ (EBD)….. ( vertical angles and bisector)
    So ∆ (ABD)= ∆EBD…. (Case SAS) => ∠ (BED)= ∠ (BAD)= alpha
    In Triangle ECD, external angle ∠ACE= 2alpha= ∠ (CED)+ ∠ (CDE) => ∠CDE= alpha
    Triangle ECD is isosceles => CD=BC+BE=BC+AB

  2. Locate E on CD such that CE = CB
    Let F be any point on extn of AB
    ∠CBE = ∠CEB = α
    ∠D =∠ FBD - ∠A = 3α/2 - α = α/2
    ∠EBD = ∠AEB - ∠D = α - α/2 = α/2
    ∆BCE, ∆ABE, & ∆BED are all isosceles.
    Hence CD = CE + ED = BC + BE = BC + AB

  3. let x = AB. Then BC = x/(2cosα).
    Then AB + BC = x(1 + 1/(2cosα)).
    Now consider triangle BCD. By sine law, CD = ((sin(3α/2))*(x/(2cosα)))/sin(α/2).
    Hence it is equivalent to prove (sin(3α/2)/sin(α/2) = 2cosα + 1
    By considering the double angle formula, 2cosα + 1 = 3 - 4(sin(α/2))^2
    The proof is then obvious by the triple angle formula.

  4. Extend AB from B till E so that BE = BC. Draw ED and EC.
    The triangle BCE is isosceles then the bisector BD is perpendicular to EC.
    Therefore the triangle CED is isosceles, ED = CD.
    Let the angle (CBD) = (DBE) = β. Then 2β = 3α.
    In the triangle BCD the angle (BDC) = 180 – (180 - 2α) – β = 2α – β; then the angle (EDC) = 2 (BDC) = 4α - 2β = 4α - 3α = α.
    Therefore the triangle AED is isosceles. Then AE = AB + BC = ED = CD

  5. Let X be the ex-center of ∆ABC opposite the vertex A.
    Draw XY ⊥ AB(extended) and XZ ⊥ AD
    It is well-known that
    the length of the tangent from A to the ex-circle
    = AZ = semi-perimeter of ∆ABC
    Next ∠XAD = α/2
    and ∠XDA = ∠YBD - ∠A = 3α/2 - α = α/2
    So ∆AXD is isosceles and XZ ⊥ AD
    Follows Z is the mid-point of AD
    ∴ AC + CD = AD = 2AZ = Perimeter of ∆ABC = AC + AB + BC,
    Hence CD = AB + BC

  6. The solution is uploaded to the following link:

  7. To Rengaraj

    Your sketch and the presentation is really nice.
    I wonder what software did you use to draw it .
    Please let us know.

  8. I use corelDRAW X5 for sketching, then I export the diagram as JPG to Microsoft word 2010. I have a software called MathType to make efective mathematic typing.

    Thaks Peter
    Have a nice Day.

  9. Picture:
    ∠BAC=a ∠BCA=2a
    Locate F on AC so that BF=AF
    ∴∠BAC=∠ABF=a (base angles theorem)
    ∴∠BFC=2a (exterior angles theorem)
    ∴BF=BC (converse of base angles theorem)
    locate E on the extended segment of AC so that AE=AB
    Goal: Prove CE=FD so then EF=CD=AE+AF=AB+BC
    ∵∠GBC is the exterior angle of ∠ABC

  10. Solution 739 sent by Sumith Peiris, Sri Lanka

    Find E on CD such that BE=ED
    Now < CBD=3A/2 and so <BDE=<BED=A/2 (A= alpha)

    So Tr BCE and Tr ABE are also isoceles, hence CD= CE+ED= BC+AB

    Sumith Peiris

    Sri Lanka