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    Internal bisectors and incenter of a triangle

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    1- Given triangle ABC, prove that an internal bisectors of an angle of a triangle divides the opposite sides (internally) into two segments proportional to the adjacent sides of the triangle. That is prove that DB/DC = AB/AC. (D is the point where e internal bisector of <A meets with BC)

    2- Given triangle ABC with in-center I prove that < BIC = 90 + ½ (< BAC)

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    Response is in a file called 'Geometry.doc"

    (1)

    Need to prove, BD/DC = AB / AC
    Lets use the sin rule for a triangle,
    For triangle, ABD
    BD/sin a = AB /sin (<BDA) ------------------------------- (1)
    <BDA = 180 - (b+a),
    So sin (<BDA ) = sin [180 - (b+a)] = sin (b+a)
    Substituting in (1),
    BD/sin a = AB /sin (b+a) --------------------(2)

    For triangle ADC using sin rule,

    DC /sin (a) = AC / sin (<ADC) ...

    Solution Summary

    This solution contains the mathematical proof of the following properties of a triangle:

    (a) Internal bisectors of an angle of a triangle divide the opposite sides (internally) into two segments proportional to the adjacent sides of the triangle

    (b) For a triangle ABC with incenter I, < BIC = 90 + ½ (< BAC)

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