The external bisector of ∠B and ∠C of ΔABC

(where AB and AC extended to E and F respectively) meet at point

If ∠BAC = 100°, then the measure of ∠BPC is

In $riangle mathrm{ABC}, angle mathrm{A}=x, angle mathrm{B}=y ; angle mathrm{C}=z$ In $Delta mathrm{PBC}, angle mathrm{PBC}+angle mathrm{PCB}+angle mathrm{BPC}=180^{circ}$ $Rightarrow frac{1}{2} angle mathrm{EBC}+frac{1}{2} angle mathrm{FCB}+angle mathrm{BPC}=180^{circ}$ $Rightarrow angle mathrm{EBC}+angle mathrm{FCB}+2 angle mathrm{BPC}=360^{circ}$ $Rightarrowleft(180^{circ}-y ight)+left(180^{circ}-z ight)+2 angle mathrm{BPC}=360^{circ}$ $Rightarrow 360^{circ}-(y+z)+2 angle mathrm{BPC}=360^{circ}$ $Rightarrow 2 angle mathrm{BPC}=y+z$ $Rightarrow 2 angle mathrm{BPC}=180^{circ}-x=180^{circ}-angle mathrm{BAC}$ $herefore angle mathrm{BPC}=90^{circ}-frac{1}{2} angle mathrm{BAC}=90^{circ}-50^{circ}=40^{circ}$ Alternative:

In $Delta mathrm{ABC}$,

$angle mathrm{BAC}=100^{circ}$

Let

$angle mathrm{ABC}=60^{circ}, angle mathrm{ACB}=20^{circ}$

(Sum of three angles of a triangle is 180°

$angle mathrm{EBC}=180^{circ}-angle mathrm{ABC}$

$=180^{circ}-60^{circ}=120^{circ}$

$herefore angle mathrm{EBP}=angle mathrm{PBC}=60^{circ}$

Similarly, $angle mathrm{BCP}=angle mathrm{PCF}=80^{circ}$

In $Delta mathrm{BPC}$

$left.angle mathrm{BPC}=180^{circ}-angle mathrm{PBC}+angle mathrm{PCB} ight)$

$=180^{circ}-left(60^{circ}+80^{circ} ight)$

$=40^{circ}$