# What is the angle of $\angle BPC$ in $\triangle BPC$

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 $\angle ACB$.

• what have you done, other than reproducing the sketch of the figure you were provided for the homework problem? – Namaste Mar 29 '18 at 17:43
• what kind or triangle is this? – Dr. Sonnhard Graubner Mar 29 '18 at 17:46
• Please read this answer addressing one way to ask a good question, when you don't "have a clue" how to proceed. – Namaste Mar 29 '18 at 17:48
• @MikeCocais to solve you only need to use $\sum$ angles = 180° applied to triangles ABC and PBC – gimusi Mar 29 '18 at 18:05
• @MikeCocais Please remember that you can choose an answer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/… – gimusi Apr 1 '18 at 8:45

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 $\Delta PBC$, $$\angle BPC+\angle PBC = \angle PCD$$ $$\angle BPC+B/2=90-C/2$$ $$\angle BPC=90-(B+C)/2$$ $$\angle BPC=90-(180-A)/2=A/2=20^{\circ}$$
• @amWhy but then, $B4 and$C$could be confused with the angles of$\Delta ABC$. I've edited the post though – Prathyush Poduval Mar 29 '18 at 18:12 • I've deleted my comment, given your edit. I was showing how you need to be consistent. None of the angles in this case should be represented by one letter; So using$\angle$and three letters is consistent. – Namaste Mar 29 '18 at 18:15 In$\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$• How you wrote$\widehat{BCP}$=$\frac{180-\widehat{BCA}}{2}$(or)$\widehat{PBC}$=$\frac{\widehat{ABC}}{2}$? It doesn't make any sense? – Mike Cocais Mar 29 '18 at 17:58 •$\widehat{BCP}=\widehat{BCA}+\widehat{ACP}$and$\widehat{PBC}=\frac{\widehat{ABC}}{2}$– user061703 Mar 29 '18 at 18:00 •$\widehat{ACP}=\frac{180^\circ-\widehat{BCA}}{2}$, it is easier to understand if you extend$BC$to the right, make it a ray$Bx$. – user061703 Mar 29 '18 at 18:01 HINT Let us indicate with$b$the angle in$B$and with$c$the angle in$C$for$\triangle ABC$. Then •$b+c+40=180 \implies c=140°-b$and •$\angle PBC = b/2$•$\angle PCB = c+(180°-c)/2=90°+c/2=160°-b/2$•$\angle BPC=180°-\angle PBC -\angle PCB=180°-b/2-160+b/2=20°$Let$\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 triangleBCP\$: $$\measuredangle BPC+\underbrace{\alpha}_{\measuredangle CBP}+\underbrace{140-2\alpha}_{\measuredangle ACB}+\underbrace{20+\alpha}_{\measuredangle ACP}=180 \Rightarrow \measuredangle BCP=20.$$