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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 $\angle BPC$?

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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 $\angle ACB$.

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    $\begingroup$ what have you done, other than reproducing the sketch of the figure you were provided for the homework problem? $\endgroup$
    – amWhy
    Mar 29, 2018 at 17:43
  • $\begingroup$ what kind or triangle is this? $\endgroup$
    – Dr. Sonnhard Graubner
    Mar 29, 2018 at 17:46
  • 2
    $\begingroup$ Please read this answer addressing one way to ask a good question, when you don't "have a clue" how to proceed. $\endgroup$
    – amWhy
    Mar 29, 2018 at 17:48
  • $\begingroup$ @MikeCocais to solve you only need to use $\sum$ angles = 180° applied to triangles ABC and PBC $\endgroup$
    – user
    Mar 29, 2018 at 18:05
  • $\begingroup$ @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/… $\endgroup$
    – user
    Apr 1, 2018 at 8:45

4 Answers 4

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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 $\Delta PBC$, $$\angle BPC+\angle PBC = \angle PCD$$ $$\angle BPC+B/2=90-C/2$$ $$\angle BPC=90-(B+C)/2$$ $$\angle BPC=90-(180-A)/2=A/2=20^{\circ}$$

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  • $\begingroup$ @amWhy isn't it standard notation to denote the angles of trangles by the symbols of it's vertices where there isn't any conflict? $\endgroup$
    – Jim Haddocc
    Mar 29, 2018 at 18:10
  • $\begingroup$ @amWhy but then, $B4 and $C$ could be confused with the angles of $\Delta ABC$. I've edited the post though $\endgroup$
    – Jim Haddocc
    Mar 29, 2018 at 18:12
  • $\begingroup$ 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. $\endgroup$
    – amWhy
    Mar 29, 2018 at 18:15
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In $\Delta{BPC}$, $\widehat{BCP}+\widehat{BPC}+\widehat{PBC}=180^\circ$, so:

$\widehat{BPC}=180^\circ-(\widehat{BCP}+\widehat{PBC})$

$=180^\circ-(\widehat{BCA}+\frac{180-\widehat{BCA}}{2}+\frac{\widehat{ABC}}{2})$

$=180^\circ-\frac{2\widehat{BCA}+180^\circ-\widehat{BCA}+\widehat{ABC}}{2}$

$=180^\circ-\frac{180^\circ+\widehat{BCA}+\widehat{ABC}}{2}$

$=180^\circ-\frac{180^\circ+180^\circ-\widehat{BAC}}{2}$

$=180^\circ-\frac{180^\circ+180^\circ-40^\circ}{2}$

$=20^\circ$

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  • $\begingroup$ How you wrote $\widehat{BCP}$ = $\frac{180-\widehat{BCA}}{2}$ (or) $\widehat{PBC}$ = $\frac{\widehat{ABC}}{2}$ ? It doesn't make any sense? $\endgroup$
    – Mike Cocais
    Mar 29, 2018 at 17:58
  • $\begingroup$ $\widehat{BCP}=\widehat{BCA}+\widehat{ACP}$ and $\widehat{PBC}=\frac{\widehat{ABC}}{2}$ $\endgroup$
    – user061703
    Mar 29, 2018 at 18:00
  • $\begingroup$ $\widehat{ACP}=\frac{180^\circ-\widehat{BCA}}{2}$, it is easier to understand if you extend $BC$ to the right, make it a ray $Bx$. $\endgroup$
    – user061703
    Mar 29, 2018 at 18:01
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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

  • $\angle PBC = b/2$
  • $\angle PCB = c+(180°-c)/2=90°+c/2=160°-b/2$
  • $\angle BPC=180°-\angle PBC -\angle PCB=180°-b/2-160+b/2=20°$
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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 $BCP$: $$\measuredangle BPC+\underbrace{\alpha}_{\measuredangle CBP}+\underbrace{140-2\alpha}_{\measuredangle ACB}+\underbrace{20+\alpha}_{\measuredangle ACP}=180 \Rightarrow \measuredangle BCP=20.$$

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