Triangle Point Conjecture: Proving $\angle BPC+\angle BAC=180^\circ$ | POTW #209

[anemone]

Here is this week's POTW:

-----Let $ABC$ be a triangle with $AC$ is longer than $AB$. The point $X$ lies on the side $BA$ extended through $A$, and the point $Y$ lies on the side $CA$ in such a way that $BX=CA$ and $CY=BA$. The line $XY$ meets the perpendicular bisector of the side $BC$ at $P$. Show that $\angle BPC+\angle BAC=180^\circ$.

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[anemone]

No one answered last week problem. :(

You can see the proposed solution as follows:

Spoiler

View attachment 5485

Let the bisector of $\angle A$ meets $BC$ at $D$ ad extend $XY$ to meet $BC$ at $E$.

Triangle $XAY$ is isosceles since $AX=BX-BA=CA-CY=AY$, hence $\angle AXY=\angle AYX$. The exterior angle of triangle $XAY$ at $A$ is $\angle A$, so $\angle A=2\angle AXY \implies \angle AXY=\angle BAD$. This suggests $AD$ is parallel to $XY$ and therefore we have

\(\displaystyle \frac{DE}{BD}=\frac{AX}{AB}\) (1)

Since $AD$ bisects $\angle A$, the following is also true:

\(\displaystyle \frac{DC}{BD}=\frac{AC}{AB}\) from which we then obtain:

\(\displaystyle \frac{DC-BD}{BD}=\frac{AC-AB}{AB}=\frac{AC-CY}{AB}=\frac{AY}{AB}=\frac{AX}{AB}\) (2)

From (1) and (2) we have

$DE=DC-BD$

But $DE=DC-EC$ so we can deduce $BD=EC$.

Let $M$ be the midpoint of $BC$. Then $M$ is also the midpoint of $DE$ and triangle $PDE$ is isosceles.

$\angle PED=\angle ECY+\angle CYE=\angle BCA+\angle AYX=\angle C+\frac{\angle A}{2}=\angle PDE$

Consequently,

$\angle DPE=\pi-2\angle C-\angle A=180^\circ-\angle C-\angle A-\angle C=\angle B-\angle C$ and so \(\displaystyle \angle DPM=\frac{\angle B-\angle C}{2}\)

Then

\(\displaystyle \tan \angle DPM=\frac{DM}{PM}\)

\(\displaystyle \begin{align*}PM&=DM \cot \angle DPM=(BM-BD)\cot \left(\frac{\angle B-\angle C}{2}\right)\\&=\left(\frac{a}{2}-\frac{ac}{b+c}\right)\cot \left(\frac{\angle B-\angle C}{2}\right)\\&=\frac{a(b-c)}{2(b+c)}\cot \left(\frac{\angle B-\angle C}{2}\right)\end{align*}\)

where $a,\,b$ and $c$ are the sides of triangle $ABC$.

We then make the substitution \(\displaystyle \frac{b-c}{b+c}=\frac{\tan \left(\frac{\angle B-\angle C}{2}\right)}{\tan \left(\frac{\angle B+\angle C}{2}\right)}\) to get

\(\displaystyle PM=\frac{a}{2\tan \left(\frac{\angle B+\angle C}{2}\right)}\)

or \(\displaystyle \tan \left(\frac{\angle B+\angle C}{2}\right)=\frac{\frac{a}{2}}{PM}=\frac{BM}{PM}\)

Also, \(\displaystyle \tan \angle BPM=\frac{BM}{PM}\) so

\(\displaystyle \tan \left(\frac{\angle B+\angle C}{2}\right)=\tan \angle BPM=\tan \frac{\angle BPC}{2}\)

From this it follows that \(\displaystyle \angle BPC=\angle B+\angle C\) and therefore we have \(\displaystyle \angle BPC+\angle BAC=\angle B+\angle C+\angle A=180^\circ\)