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

Posted by Antonio Gutierrez at 12:27 PM
Labels: , , , ,

6 comments:

  1. AjitJune 14, 2009 at 10:06 PM

    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

    ReplyDelete
  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.

    ReplyDelete
  3. AnonymousOctober 15, 2009 at 2:08 PM

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

    ReplyDelete
  4. AnonymousNovember 20, 2009 at 12:23 PM

    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

    ReplyDelete
  5. AnonymousDecember 30, 2009 at 5:34 AM

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

    ReplyDelete
  6. Sumith PeirisJuly 5, 2023 at 5:33 AM

    Complete the square APDQ, P on AB and Q on AC

    S(ABC) = bc/2 = (c/2)x/V2 + (b/2)x/V2

    So V2/x = 1/b + 1/c

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

Subscribe to: Post Comments (Atom)