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