$P$ is a point on the angular bisector of $\angle A$. Show that $\frac{1}{AB}+\frac{1}{AC}$ doesn't depend on the line through $P$ 
The point $P$ is on the angular bisector of a given angle $\angle A$. A line $L$ is drawn through $P$ which intersects with the legs of the angle in $B$ and $C$. Show that $$\dfrac{1}{AB} + \dfrac{1}{AC}$$ is not dependent on the choice of the line $L$.

I started by drawing it up and came to the conclusion that $AC$ + $AB$ should always be the same because of the angles where $\triangle$ $APB$ and $\triangle$ $APC$ are. So I wrote $\angle APB$ = $\beta$ and $\angle APC$ = $\theta$ 
Then I wrote $L*\sin\beta$ = $AB$ and $L*\sin\theta$ = $AC$. Then I added them so: $L(\sin(\alpha + \theta)) = AB + AC$. But $\sin (\alpha + \theta)$ should be $0$ since $\alpha + \theta$ is $180 ^{\circ}$, and $AB + AC$ is not $0$
 A: Here is a purely geometric proof.  I assume that $AP$ is an internal angular bisector of $\angle A$.  
Let $C'$ be the image of $C$ under the reflection about the line $AP$.  Let $D$ be the point on the ray $AB$ such that $\angle APD=90^\circ$.  Extend $DP$ to meet the ray $AC$ at $E$.  

We have $\triangle CPE\cong \triangle C'PD$.   Therefore,
$$\angle C'PD=\angle CPE=\angle DPB.$$
So, $PD$ is the angular bisector of $\angle BPC'$.  As $\angle APD=90^\circ$, $A$ and $D$ harmonically divide $B$ and $C'$.  In other words, when we include signs for oriented lengths, we have
$$\frac{BD}{DC'}=-\frac{BA}{AC'}.$$
Hence,
$$\frac{BD}{AB}=-\frac{BD}{BA}=\frac{DC'}{AC'}.$$
Thus,
$$\frac{AD}{AB}=\frac{AB+BD}{AB}=1+\frac{BD}{AB}=1+\frac{DC'}{AC'}=2-\left(1-\frac{DC'}{AC'}\right).$$
That is,
$$\frac{AD}{AB}=2-\frac{AC'-DC'}{AC'}=2-\frac{AC'+C'D}{AC'}=2-\frac{AD}{AC'}.$$
This means
$$\frac{1}{AB}+\frac{1}{AC}=\frac{1}{AB}+\frac{1}{AC'}=\frac{1}{AD}\left(\frac{AD}{AB}+\frac{AD}{AC'}\right)=\frac{2}{AD}.$$
Because $A$ and $D$ are fixed, $\frac{1}{AB}+\frac{1}{AC}$ is independent of the choice of $L$.

On the other hand, if $AP$ is an external angular bisector of $\angle A$, we can modify the proof above slightly.  It can be proven that $$\frac{1}{AB}-\frac{1}{AC}=\frac{2}{AD},$$
with the point $D$ as defined above.  Here, $B$ is assumed to lie between $P$ and $C$.
A: Pure synthetic solution:
Let $(XYZ)$ denot an area of triangle $XYZ$. We see that: $$(BAC) = (BAP)+(CAP)$$
So $${AB\cdot AC \cdot \sin (2\alpha)\over 2}={AB\cdot AP \cdot \sin (\alpha)\over 2}+{AP\cdot AC \cdot \sin (\alpha)\over 2}$$
If we divide this by ${AB\cdot AC\cdot AP\cdot \sin \alpha\over 2}$ we get $$ {2\cos \alpha \over AP} = {1\over AC}+{1\over AB}$$
and we are done!
A: Since $P$ is on the angle bisector of $\angle A$, we have
$$
\sin\angle PAB=\sin\angle PAC
$$
Since $BPC$ is the line $L$, $\angle APB$ and $\angle APC$ are supplementary, hence
$$
\sin\angle APB=\sin\angle APC
$$
Dividing and applying the sine rule,
$$
\frac{BP}{AB}=\frac{\sin\angle BAP}{\sin\angle APB}=\frac{\sin\angle CAP}{\sin\angle APC}=\frac{CP}{AC}
$$
(this is the angle bisector theorem: the (internal) angle bisector of a triangle cuts the opposite side in proportion to the side lengths.)
So again with the sine rule,
$$
\frac{1}{AB}+\frac{1}{AC}=\frac{BC}{AC}\times\frac{1}{BP}=\frac{\sin\angle A}{BP\sin\angle ABC}=\frac{\sin\angle A}{AP\sin\angle PAB}
$$
is independent of $L$.
A: In the standard notations by low of sines we obtain:
$$\frac{1}{AB}+\frac{1}{AC}=\frac{1}{AP}\left(\frac{AP}{AB}+\frac{AP}{AC}\right)=\frac{1}{AP}\left(\frac{\sin\beta}{\sin\left(\frac{\alpha}{2}+\beta\right)}+\frac{\sin\gamma}{\sin\left(\frac{\alpha}{2}+\beta\right)}\right)=$$
$$=\frac{2\sin\frac{\beta+\gamma}{2}\cos\frac{\beta-\gamma}{2}}{AP\sin\left(\frac{180^{\circ}-\beta-\gamma}{2}+\beta\right)}=\frac{2\cos\frac{\alpha}{2}}{AP},$$ which does not depend on $L.$
