Tuesday, November 4, 2014

Geometry Problem 1055: Triangle, Altitude, Perpendicular, Area, Circumradius, Circle

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 1055: Triangle, Altitude, Perpendicular, Area, Circumradius, Circle.

3 comments:

  1. Let S be the area. Let BC=a, AC=b, AB=c.

    Since ΔABC~ΔDBE, thus DE/AC=BD/AB.
    d / b = (c sinA sinC) / c
    d = b sinA sinC = b (a/2R) (c/2R)

    Hence, S = abc/4R = Rd.

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  2. D,F,E,B are concyclic and they lie on the circle on BF as diameter.
    In this circle d = chord EF = (diameter BF) x sin B
    (i.e.) d = h Sin B where h denotes height BF.
    In ΔABC we have
    2Δ= h.b = h. 2R sin B = 2R .h Sin B = 2R.d
    Hence Δ = R.d

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  3. R=BC/2*BA/BF
    d=AC*BD/BA
    Rd=BC*AC*BD/(2BF)
    Due to similar triangles BC*BD=BF^2
    Therefore, Rd=AC*BF/2=[ABC]

    ReplyDelete