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Given $\Delta$$ABC$ with $AB$$\bot$$BC$, $BD$ is the altitude to $AC$, $AF$ bisects angle $BAC$, $AP=12$ , and $PF=8$. Find $\tan(\angle BAF$) and find $PD$.

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Efforts made: I've tried to apply angle bisector theorem but i couldn't get anything usefull ,and i've also tried to add some lines such as $FD$ but it didn't help me much though.

Note: I don't want the solution to the problem,only hints. Thanks in advance.

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Let $BC=a,CA=b,AB=c$.

I don't want the solution to the problem,only hints

I think you can get the answers using the followings :

  • Since $AF$ bisects angle $BAC$, we have $AB:AC=BF:FC$ from which we can represent $BF,FC$ by $a,b,c$.

  • The area of $\triangle{ABC}$ equals $\frac 12\times AB\times BC=\frac 12\times BD\times AC$ from which we can represent $BD$ by $a,b,c$.

  • Since $\triangle{DAP}$ is similar to $\triangle{BAF}$, we can represent $DA$ by $c$.

  • Since $\triangle{CBA}$ is similar to $\triangle{CDB}$, using $b^2-a^2=c^2$, we can represent $c$ by $b$, and $a$ by $b$ from which we can get $\tan(\angle{BAF})$.

  • Having $\cos(\angle{BAF})$ leads to $c$ and $PD$.

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  • $\begingroup$ Thanks for the hints,one question: From your first hint how can you represent $FC$ by $a,b,c$ ? $\endgroup$ – Nameless Sep 19 '15 at 16:29
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    $\begingroup$ @Jhon: Since $c:b=BF:FC$, i.e. $c:b=a-FC:FC$, we have $FC=\frac{ab}{b+c}$. $\endgroup$ – mathlove Sep 19 '15 at 16:47
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HINTS:

Length of angle bisector at angle $ $A$ = \dfrac {2 b c }{ b+c} \cos A/2, $ with opposite sides $BD$ and $ BC$,

Similarity of triangles within a right angled triangle with results like $ AB^2 = AD\cdot AC, $

and, algebraic manipulations.

RESULTS:

$ \tan BAF = \frac12 $

$ PD\approx 5.36656.$

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