# $P$ 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 $$0$$

• Can you post the question in your native language? Maybe someone else can help translate. For now, I don't understand the question. Commented Nov 5, 2018 at 9:46
• @Batominovski I'm sorry! Is this better? Commented Nov 5, 2018 at 9:52

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 $$C$$.

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!

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 $$L$$.

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 $$L.$$