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 $0$
Best Answer
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 $C$.