Saturday, December 26, 2015

Geometry Problem 1174: Triangle, Quadrilateral, Double, Triple, Angle, Congruence, Excenter, Angle Bisector

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

Click the figure below to view more details of problem 1174.

Online Math: Geometry Problem 1174: Triangle, Quadrilateral, Double, Triple, Angle, Congruence, Excenter, Angle Bisector.

2 comments:

  1. Let the bisector of < ABC meet AC at E, CD at F and AD at G.
    For ease of typing I use @ for alpha and $ for theta

    Easily < ABO = OBC = @/2 and < CBE = @ so that < EBD = 2@
    Also since BG is the perpendicular bisector of AD, < BAE = EAF = BDE = EDF = $
    So C is the incentre of Tr. ABF so < DFG = < AFG = < AFC = < CFE and each of these must therefore be essentially 60 degrees since the last 3 of these 4 angles add upto 180.
    Hence from Tr. BDF, 2(&+$) = 60 and so
    &+$ = 30.....(1)

    In Tr. ABC, OA is an external angle bisector at A so x+@/2 = 90 -$/2 from which
    x + (@+$)/2 = 90
    Substituting @+& = 30 from (1) therefore
    x+30/2 = 90 and hence x = 75

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  2. http://s17.postimg.org/59873mej3/pro_1174.png

    Let angle bisector of angle ABD meet CD at F and AD at E
    Observe that due to symmetry we have
    ∠(BAF)= ∠ (BDF)=2. Theta
    And tri. AFD is isosceles
    AC and BC are angles bisectors of ∠ (BAC) and ∠ (CFA)
    So ∠ (BFC)= ∠ (CFA)=2.alpha+2.theta => ∠ (FDA)=alpha+theta
    In right tri. BED we have alpha+theta=30
    We have∠ (HOA)= theta/2 and ∠ (ABO)=alpha/2
    In right tri. HBO we have x+theta/2+alpha/2= 90 => x= 75

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