Wednesday, August 18, 2010

Problem 505: Right Triangle, Cevian, Sum of Segments, Angles

Geometry Problem
Click the figure below to see the complete problem 505 about Right Triangle, Cevian, Sum of Segments, Angles.

Problem 505. Right Triangle, Cevian, Sum of Segments, Angles


See also:
Complete Problem 505

Level: High School, SAT Prep, College geometry

8 comments:

  1. With the given data is it possible to prove that BD bisects /_B ? Easy to do this by trigonometry but how do we do it by plane geometry alone?
    Of course, this makes x=15 deg.
    Vihaan

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  2. draw DBM = x => BM = a, get AH = a + d , H on AC
    => CH = DM => HM = d => MA = a => MBA = 2x, B = 6x
    => x = 15°

    ReplyDelete
  3. To c.t.e.o

    It not clear to me why CH=DM in your solution. Please explain.

    Peter Tran

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  4. geting AH = a + d give us ang HBC = x
    drawing DBM = x => ▲DMB = ▲HBC
    (BM = BC = a, MCB = CMB, MBD = CBH = x : ASA )
    In triangle MBC altitude is bisector, giving BM = a

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  5. To: c.t.o.e.
    What a nice soluion, well done!
    Ajit

    ReplyDelete
  6. To Joe
    Thanks

    If BG is altitude of ABC than BG is median of MBC,DBH
    => MD = HC as diferences of equal parts

    ( I liked your solution of P 167 )

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  7. To c.t.e.o
    Your solution is the most beautiful solution I've ever seen.

    ReplyDelete
  8. Trisect < CBD by BE and BF, E and F on AC and BE thus being an altitude as well

    Let CE = e so that DE = EF = d-e and CF = 2e-d

    < AFB = 90-x = < ABF so AB = AF = a+d = b - (2e-d) hence e = (b-a)/2......(1)

    Now CB is tangential to Tr. ABE at B so,

    a^2 = eb

    Substituting for e in (1),

    b(b-a)/2 = a^2 hence b^2 - ab - 2b^2 = 0

    Solving the quadratic and rejecting the negative solution we have b = 2a hence 2x = 30 and x = 15

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete