Sunday, November 8, 2020

Geometry Problem 1482: Right Triangle, Perpendicular, Double Angle, Measurement

Geometry Problem 1482. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

Geometry Problem 1482: Right Triangle, Perpendicular, Double Angle, Measurement.

6 comments:

  1. Draw a)EH perpend AC b)AP bisector of angEAC c)EMT perpend AP
    => Tr DET isosceles, from EM=6 => ET=DE=12

    ReplyDelete
  2. https://photos.app.goo.gl/wrmr2bHVba98ZA3k8

    Draw angle bisector AH of angle CAE
    Draw EF ⊥AH ( see sketch)
    Note that ∠ (CEF)= 2. Alpha and triangle EAF is isosceles
    Triangle ABE congruent to AEH and AFH
    So EH=HF=BE=6 so EF=12
    In triangle EFC , external angle ∠ (DFE)= theta+2. Alpha
    In triangle AED, external angle ∠ (EDF)= theta+2. Alpha
    So tringle DEF is isosceles and DE=EF=12

    ReplyDelete
  3. Let:
    AB = a

    1.)
    3α + Θ +90 = 180 => Θ = 90 - 3α

    2.)
    Angle x / (2sin(α)) = √(a² + 36)
    _________________

    From (3) and (4):

    x = 12.

    ReplyDelete
  4. Extend EB to F such that BE = BF = 6 so that BA bisects < FAE

    Then < FEA = 2.Alpha + Theta = 90 - Alpha from which easily < CED = 4.Alpha = < CAF

    Hence ADEF is concyclic and since EA bisects < CAF, it follows that DE = EF = 12

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. Fold Tr.ABE along AE such that B' is the reflection of B (ABEB' is cyclic )
    m(BAE)=m(EAB')=Alpha=>AB' is angle bisector of m(EAD). Extend EB' to meet AC at F
    Since AB' bisects m(EAF) and AB' perpendicular to EF => EF=2.EB'=2.BE=12
    m(EFA)=180-m(EFC)=2Alpha+Theta=m(EDF)
    => EDF is an isosceles triangle and ED=EF=12

    ReplyDelete