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Math Help - trigo(4)

  1. #1
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    trigo(4)

    A triangle ABC is right angled at A . The angle bisector of angle A meets BC at D . If BC = a , CA =b , and AB=c , prove that

    c/b = BD/DC

    i just couldn't get the sketch right .

    The line that meets BC at D is not perpendicular right ?
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  2. #2
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    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    A triangle ABC is right angled at A . The angle bisector of angle A meets BC at D . If BC = a , CA =b , and AB=c , prove that

    c/b = BD/DC

    i just couldn't get the sketch right .

    The line that meets BC at D is not perpendicular right ?
    Correct! AD is not perpendicular to BC.

    \angle BAD = \angle DAC = 45^o

    Suppose the length of AD is l.

    Then, using the \tfrac12ab\sin C formula, write down the areas of \triangle's ABD and ACD. (Hint: the area of \triangle ABD = \tfrac12cl\sin 45^o.)

    Then consider these two triangles as though their bases lay along the line BDC. So they have a common height. So the ratio of their areas is the same as the ratio of the lengths of their bases which is C" alt="BDC" />.

    Then use your first two areas, and you're there.

    (Incidentally, although you're told that \angle A = 90^o, it doesn't have to be, provided it is bisected by AD.)

    Grandad
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  3. #3
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    What are you trying to demonstrate is the angle bisector theorem. Look at Angle bisector theorem - Wikipedia, the free encyclopedia , there is a demonstration for the general case, but I hope you can adapt it. AD\perp BC only if the triangle is isosceles.
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  4. #4
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    Quote Originally Posted by thereddevils View Post
    A triangle ABC is right angled at A . The angle bisector of angle A meets BC at D . If BC = a , CA =b , and AB=c , prove that

    c/b = BD/DC

    i just couldn't get the sketch right .

    The line that meets BC at D is not perpendicular right ?
    This is the continuation of this question which i cant figure out .

    Prove also taht

     <br />
AD=\frac{\sqrt{2}a \sin B}{1+ \tan B}<br />


    Thanks
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  5. #5
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    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    This is the continuation of this question which i cant figure out .

    Prove also taht

     <br />
AD=\frac{\sqrt{2}a \sin B}{1+ \tan B}<br />


    Thanks
    Using the fact that the ratio C=c:b" alt="BDC=c:b" /> and BD+DC = BC =a, we can say that

    BD = \frac{c}{b+c}\times BC=\frac{ca}{b+c}

    Now use the Sine Rule on \triangle ABD:

    \frac{AD}{\sin B}= \frac{BD}{\sin 45^o} = \sqrt2BD, since \sin45^o = \frac{1}{\sqrt2}

    \Rightarrow AD = \frac{\sqrt2ca\sin B}{b+c}= \frac{\sqrt2\sin B}{1+\frac{b}{c}}=\frac{\sqrt2\sin B}{1+\tan B}

    Grandad
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