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Thread: 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! $\displaystyle AD$ is not perpendicular to $\displaystyle BC$.

    $\displaystyle \angle BAD = \angle DAC = 45^o$

    Suppose the length of $\displaystyle AD$ is $\displaystyle l$.

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

    Then consider these two triangles as though their bases lay along the line $\displaystyle BDC$. 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 $\displaystyle BDC$.

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

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

    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. $\displaystyle AD\perp BC$ 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

    $\displaystyle
    AD=\frac{\sqrt{2}a \sin B}{1+ \tan B}
    $


    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

    $\displaystyle
    AD=\frac{\sqrt{2}a \sin B}{1+ \tan B}
    $


    Thanks
    Using the fact that the ratio $\displaystyle BDC=c:b$ and $\displaystyle BD+DC = BC =a$, we can say that

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

    Now use the Sine Rule on $\displaystyle \triangle ABD$:

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

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

    Grandad
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