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In $\triangle ABC$, the internal bisector of $\angle ABC$ and the external bisector of $\angle ACB$ meet at $P$. If $\angle BAC = 40^\circ$ what is the measure of $\angle BPC$?
My try: i) Sum of angles of a triangle is $180^\circ$.
ii) Vertical opposite angles are equal.
We need to find $\angle BPC$. By i) we know $\angle BPC = 180^\circ - \angle PCA - \angle PKC$. So the line pass through points $P$ and $C$ is perpendicular to internal bisector of $\angle ACB$.
Extend $BC$ to $D$, so that $\angle ACD$ is an exterior angle of the triangle. Thus, $$\angle PCD = \angle ACD/2= 90-C/2$$
Using the exterior angle sum property in $\Delta PBC$, $$\angle BPC+\angle PBC = \angle PCD$$ $$\angle BPC+B/2=90-C/2$$ $$\angle BPC=90-(B+C)/2$$ $$\angle BPC=90-(180-A)/2=A/2=20^{\circ}$$
In $\Delta{BPC}$, $\widehat{BCP}+\widehat{BPC}+\widehat{PBC}=180^\circ$, so:
$\widehat{BPC}=180^\circ-(\widehat{BCP}+\widehat{PBC})$
$=180^\circ-(\widehat{BCA}+\frac{180-\widehat{BCA}}{2}+\frac{\widehat{ABC}}{2})$
$=180^\circ-\frac{2\widehat{BCA}+180^\circ-\widehat{BCA}+\widehat{ABC}}{2}$
$=180^\circ-\frac{180^\circ+\widehat{BCA}+\widehat{ABC}}{2}$
$=180^\circ-\frac{180^\circ+180^\circ-\widehat{BAC}}{2}$
$=180^\circ-\frac{180^\circ+180^\circ-40^\circ}{2}$
$=20^\circ$
HINT
Let us indicate with $b$ the angle in $B$ and with $c$ the angle in $C$ for $\triangle ABC$. Then
and
Let $\alpha=\measuredangle ABP=\measuredangle CBP$. Then: $$\begin{align}&\measuredangle ACB=180-40-2\alpha=140-2\alpha \\ &\measuredangle ACP=\frac12(180-\measuredangle ACB)=\frac12(180-(140-2\alpha))=20+2\alpha.\end{align}$$ Now the sum of angles of triangle $BCP$: $$\measuredangle BPC+\underbrace{\alpha}_{\measuredangle CBP}+\underbrace{140-2\alpha}_{\measuredangle ACB}+\underbrace{20+\alpha}_{\measuredangle ACP}=180 \Rightarrow \measuredangle BCP=20.$$
