Sunday, June 14, 2009

Problem 303. Triangle, Angle bisector of 90 degrees

Proposed Problem
Click the figure below to see the complete problem 303 about Triangle, Angle bisector of 90 degrees.

 Problem 303: Triangle, Angle bisector of 90 degrees.
See also:
Complete Problem 303
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

5 comments:

  1. By sine rule in Tr. DAC, x/b = sin(C)/sin(B+45). Similarly, x/c = sin(B)/sin(C+45) = sin(B)/sin(B+45) since (B+45) and (C+45) are supplementary.
    Thus, x/b + x/c = [sin(B) + sin(C)]/sin(B+45)=[(c/a+b/a)/sin(B+45)]= (b+c)/(asinB/V2 +acos(B)/V2) = V2(b+c)/(b+c) = V2 or 1/b + 1/c = V2/x
    Ajit: ajitathle@gmail.com

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  2. Mrudul M. ThatteJune 15, 2009 at 6:55 AM

    Let BC = a
    Case 1) Triangle ABC is scelene.
    angle BAD = 45
    So,
    (BD)*(BD) = c*c + x*x - (square root of 2)*c*x
    angle CAD = 45
    So,
    (CD)*(CD) = b*b + x*x - (square root of 2)*b*x

    As AD is angle bisector of angle BAC,
    BD/CD =c/b
    So, [(BD)*(BD)]/[(CD)*(CD)] = [c*c] / [b*b]

    By substituting value of (BD)*(BD) and (CD)*(CD) in this equation we get,
    [c*c + x*x - (square root of 2)*c*x] /
    [b*b + x*x - (square root of 2)*b*x]
    = [c*c] / [b*b]
    By cross multiplication we get,
    b*b*c*c + b*b*x*x - (square root of 2)b*b*c*x
    = b*b*c*c + c*c*x*x - (square root of 2)b*c*c*x

    So,
    b*b*x*x - (square root of 2)b*b*c*x
    =c*c*x*x - (square root of 2)b*c*c*x

    By dividing this equation by x,

    b*b*x - (square root of 2)b*b*c
    =c*c*x - (square root of 2)b*c*c

    So,
    b*b*x - c*c*x
    = (square root of 2)*b*b*c
    - (square root of 2)*b*c*c

    So,
    x*(b+c)(b-c) = (square root of 2)*b*c*(b-c)
    As (b-c) is nonzero number (Reason : Case 1),
    x*(b+c) = (square root of 2)*b*c
    So,
    x/(square root of 2) = (b*c)/ (b+c)
    So,
    (square root of 2)/x = (b+c)/(b*c)
    = b/(b*c) + c/(b*c)
    = (1/c) + (1/b)

    Case 2) b = c or triangle ABC is isosceles triangle.

    According to properties of an right isosceles triangle,
    x = a/2

    1/b + 1/c = 2/b

    But b = a/(Square root of 2)
    So,
    1/b + 1/c = 2/[a/(Square root of 2)]
    = 2 * (Square root of 2) / a
    = (Square root of 2)/x


    Hence the proof.

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  3. http://img379.imageshack.us/img379/3304/coz.gif

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  4. Draw DE perpendicular to AC, => ADE has DE=AE
    =>x'2=DE'2+DE'2 or x'2=2DE'2
    ABC and DEC are similar (AB//DE)
    =>DE/c = b-DE/b => DE = cb / c+b

    substitute DE => x'2 = 2 cb / c+b '2
    give the result

    P.S. '2 mean exponent 2

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  5. [ABC]=[ACD]+[ADB]
    then
    bc=bxsin45+cxsin45
    dividing both sides by bcxsin45 gives the solution.
    .-.

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