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If ABC is a triangle with AB=20,BC=22 and CA=24. Let D lie on BC such that AD is the angle bisector of ∠BAC. What is AD2?

  • maths -

    I assume the last sentence is
    "What is AD?" since AD2 does not mean much.

    Hints:

    Use cosine rule to find ∠A, in triangle ABC.
    A=arccos((20²+24²-22²)/(2*20*24))
    =arccos(41/80)
    = 59° (approx.)

    Then
    Area of triangle ABC
    =(1/2)AB.AC.sin(A)
    =(1/2)*20*24*(11√39)/80
    =33√39

    Since AD is angle bisector, the altitudes of D to AB equals that to AC.
    Therefore AD divides the area of triangle ABC in ratios of the bases AB and AC. Thus
    Area of ΔADC=(33√39)*(24/(20+24))
    =18√39

    Area of ΔADC can also be found by the trigonometric formula
    Area of ΔADC
    =AD*AC*sin(A/2) ... since AD bisects A

    =>
    18√39 = AD * 24 *sin(A/2)
    =>
    AD=3√39/2sin(A/2)=19 (approx.)

  • maths -

    It is asking for the length of AD squared or AD^2.

  • maths -

    The solution is given for AD.
    Are you able to calculate AD²?

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