Wednesday, April 29, 2015

Geometry Problem 1114: Right Triangle, Angle Trisection, Isosceles, Congruence

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

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1114: Right Triangle, Angle Trisection, Isosceles, Congruence.


  1. In ΔACC₁,
    AC₁/sin(2C/3) = AC/sin(90°+C/3)
    AC₁ = AC sin(2C/3)/cos(C/3) = 2 AC sin(C/3)

    In ΔACB₂,
    AB₂/sin(C/3) = AC/sin(A/3+C/3) = 2 AC
    AB₂ = 2 AC sin(C/3)

    Thus, AC₁ = AB₂.
    Similarly, CA₁ = CB₂.

    So AA₁ is the perpendicular bisector of C₁B₂,
    thus AC₁A₁B₂ is a kite, therefore A₁C₁=A₁B₂.

    Similarly, C₁A₁=C₁B₂.
    Hence, ΔB₂A₁C₁ is equilateral and so B₂A₁=B₂C₁.


    1. Draw line CEF ⊥AA2 ( E on AA2 extension, F on AA1 extension)
    Since ∠A+ ∠ C= 90 => ∠ (AB2C)= 180- 1/3(A+C)= 150
    In triangle CAF since AE is an altitude and angle bisector so CAF is isosceles.
    We have ∠ (BCE)= ∠ (BAE)= 2/3A= ∠ (CAF)
    Triangle CAF similar to FCA1 ( case AA) => CFA1 isosceles => CA1=CF
    Triangle CEB2 is 30-60-90 triangle so CB2= 2 CE= CF
    So CA1=CF=CB2.
    Similarly we will have AC1= AH= AB2.
    2. Since AA1 and CC1 are perpendicular bisectors of C1B2 and A1B2 so A1B2C1 is a equilateral triangle
    So A1B2=C1B2

  3. < A + < C = 90. So < AB1C in Tr. AB1C is 120. But B1B2 bisects this angle. Hence < C1B1A = AB1B2= B2B1C = < A1B1C = 60.

    So Tr.s AC1B1 & AB2B1 are congruent (SAA) and so AC1 = AB2 and B1B2 = B1C1

    Similarly CA1 = CB2

    Also C1B1=B1B2=B1A1, hence C1B2 = A1B2

    Sumith Peiris
    Sri Lanka